{"id":18,"date":"2023-09-17T11:09:13","date_gmt":"2023-09-17T11:09:13","guid":{"rendered":"https:\/\/mathority.org\/pt\/zero-entre-zero-0-0-indeterminacao\/"},"modified":"2023-09-17T11:09:13","modified_gmt":"2023-09-17T11:09:13","slug":"zero-entre-zero-0-0-indeterminacao","status":"publish","type":"post","link":"https:\/\/mathority.org\/pt\/zero-entre-zero-0-0-indeterminacao\/","title":{"rendered":"Indetermina\u00e7\u00e3o zero entre zero (0\/0)"},"content":{"rendered":"<p>Neste artigo explicamos como salvar o limite de uma fun\u00e7\u00e3o quando ela d\u00e1 incerteza 0\/0. Al\u00e9m disso, voc\u00ea poder\u00e1 praticar com exerc\u00edcios resolvidos sobre a indetermina\u00e7\u00e3o de zero entre zero. <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/indetermination-zero-entre-zero-00.webp\" alt=\"\" class=\"wp-image-888\" width=\"213\" height=\"214\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"como-resolver-la-indeterminacion-cero-entre-cero-00\"><\/span> Como resolver a indetermina\u00e7\u00e3o zero entre zero (0\/0)<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Veremos ent\u00e3o como calcular o limite de uma fun\u00e7\u00e3o quando ela d\u00e1 indetermina\u00e7\u00e3o zero entre zero (0\/0). Para fazer isso, calcularemos um exemplo passo a passo:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8756377c47addb7fc7c1a9101d6fe29c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to 2} \\frac{x^2-x-2}{x^2-3x+2}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"123\" style=\"vertical-align: -14px;\"><\/p>\n<\/p>\n<p> Primeiro tentamos calcular o limite substituindo o valor de x na fun\u00e7\u00e3o:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-57f54a6e222522bf26a65b6dee7e2334_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to 2} \\frac{x^2 - x -2}{x^2-3x+2}=\\frac{2^2 -2-2}{2^2-3\\cdot 2+2}=\\frac{0}{0}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"286\" style=\"vertical-align: -14px;\"><\/p>\n<\/p>\n<p> Mas obtemos a indetermina\u00e7\u00e3o 0 dividida por 0. <\/p>\n<div style=\"background:linear-gradient(to bottom, #FFFFFF 0%, #FFE0B2 100%); padding-top: 23px; padding-bottom: 0.5px; padding-right: 30px; padding-left: 30px; border: 2px dashed #FF9B28; border-radius:20px; margin-bottom:30px\">\n<p style=\"text-align:left\"> Quando o limite de uma fun\u00e7\u00e3o pontual d\u00e1 a <strong>incerteza 0\/0<\/strong> , \u00e9 necess\u00e1rio fatorar os polin\u00f4mios do numerador e do denominador e depois simplificar os fatores comuns.<\/p>\n<\/div>\n<p> Devemos, portanto, fatorar os polin\u00f4mios do numerador e denominador da fra\u00e7\u00e3o. Para fazer isso, usamos a regra de Ruffini: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/factorisation-indetermination-00.webp\" alt=\"fatora\u00e7\u00e3o de indetermina\u00e7\u00e3o 0\/0\" class=\"wp-image-894\" width=\"429\" height=\"312\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p> <span style=\"color:#FF9B28;\">\u27a4<\/span> Se voc\u00ea n\u00e3o sabe <u style=\"text-decoration-color:#FF9B28;\">como fatorar um polin\u00f4mio<\/u> , recomendamos que veja a explica\u00e7\u00e3o em nosso site especializado em polin\u00f4mios: <u style=\"text-decoration-color:#FF9B28;\">www.polinomios.org<\/u><\/p>\n<p> Assim, uma vez fatorados os polin\u00f4mios, o limite \u00e9 o seguinte:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-225121e089bcdcdb7ea055e9fd01c61d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to 2} \\cfrac{x^2-x-2}{x^2-3x+2}=\\lim_{x \\to 2}\\frac{(x+1)(x-2)}{(x-1)(x-2)}\" title=\"Rendered by QuickLaTeX.com\" height=\"46\" width=\"288\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p> Podemos agora simplificar o limite eliminando os fatores que se repetem no numerador e no denominador da fra\u00e7\u00e3o:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-461b8157cb8cdf50595cc35c590dc720_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to 2} \\cfrac{(x+1)\\cancel{(x-2)}}{(x-1)\\cancel{(x-2)}}=\\lim_{x \\to 2} \\cfrac{(x+1)}{(x-1)}\" title=\"Rendered by QuickLaTeX.com\" height=\"45\" width=\"254\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E finalmente, recalculamos o limite:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c1d6173c0ee113815a638c71e1c36e7a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to 2} \\cfrac{x+1}{x-1}=\\cfrac{2+1}{2-1}=\\cfrac{3}{1}=\\bm{3}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"206\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p> Como voc\u00ea pode ver, uma vez fatorados e simplificados os polin\u00f4mios, \u00e9 muito f\u00e1cil encontrar a solu\u00e7\u00e3o no limite. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"indeterminacion-00-con-raices\"><\/span>Indetermina\u00e7\u00e3o 0\/0 com ra\u00edzes<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Acabamos de ver como as indetermina\u00e7\u00f5es 0\/0 das fun\u00e7\u00f5es racionais s\u00e3o resolvidas. Por\u00e9m, se o limite for de uma fun\u00e7\u00e3o irracional (ou radical), a indetermina\u00e7\u00e3o 0\/0 \u00e9 resolvida de forma diferente.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-077724689257ed57dfb621061adba77e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to 1} \\frac{x-1}{\\sqrt{x}-1}\" title=\"Rendered by QuickLaTeX.com\" height=\"40\" width=\"89\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p> Primeiro, tentamos resolver o limite realizando as seguintes opera\u00e7\u00f5es:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-27ddc1a2cc56460b9f511d4c7d6b48c0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to 1} \\frac{x-1}{\\sqrt{x}-1}=\\frac{1-1}{\\sqrt{1}-1}=\\frac{0}{0}\" title=\"Rendered by QuickLaTeX.com\" height=\"40\" width=\"207\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p> Mas obtemos zero sobre zero indetermina\u00e7\u00e3o. <\/p>\n<div style=\"background:linear-gradient(to bottom, #FFFFFF 0%, #FFE0B2 100%); padding-top: 23px; padding-bottom: 0.5px; padding-right: 30px; padding-left: 30px; border: 2px dashed #FF9B28; border-radius:20px; margin-bottom:30px\">\n<p style=\"text-align:left\"> Se o <strong>limite de uma fun\u00e7\u00e3o com ra\u00edzes d\u00e1 indetermina\u00e7\u00e3o 0\/0<\/strong> , deve-se multiplicar o numerador e o denominador da fra\u00e7\u00e3o pelo conjugado da express\u00e3o radical.<\/p>\n<\/div>\n<p> \u27a4 Lembre-se que o conjugado \u00e9 a mesma express\u00e3o irracional, mas com o sinal do meio modificado.<\/p>\n<p> A seguir, multiplicamos o numerador e o denominador da fra\u00e7\u00e3o pelo conjugado da express\u00e3o radical:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-28c2950e1fe30fbda237ea5d154fdbd9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to 1} \\frac{\\left(x-1\\right)\\cdot\\left(\\sqrt{x}+1\\right)}{\\left(\\sqrt{x}-1\\right)\\cdot\\left(\\sqrt{x}+1\\right)}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"185\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p> Dentro deste tipo de limites, ao realizar este passo obteremos sempre uma identidade not\u00e1vel que podemos simplificar. Neste caso, no denominador temos o produto de uma soma e uma diferen\u00e7a, portanto:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c2ec8901a2ec84af3d8b70143894ca38_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to 1} \\frac{\\left(x-1\\right)\\cdot\\left(\\sqrt{x}+1\\right)}{\\left(\\sqrt{x}\\right)^2-1^2}\" title=\"Rendered by QuickLaTeX.com\" height=\"46\" width=\"170\" style=\"vertical-align: -20px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-785777714acb47d4f3772980575c4dac_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to 1} \\frac{\\left(x-1\\right)\\cdot\\left(\\sqrt{x}+1\\right)}{x-1}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"170\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p> Simplificamos o fator que se repete no numerador e no denominador:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-27813ee911be89725aa9e79230e1e76a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to 1} \\frac{\\cancel{\\left(x-1\\right)}\\cdot\\left(\\sqrt{x}+1\\right)}{\\cancel{x-1}}=\\lim_{x \\to 1}\\left(\\sqrt{x}+1\\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"296\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p> E desta forma podemos encontrar o resultado do limite: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a7746fb54d762beeeb44041650a78004_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to 1}\\left(\\sqrt{x}+1\\right)=\\sqrt{1}+1=2\" title=\"Rendered by QuickLaTeX.com\" height=\"29\" width=\"212\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"ejercicios-resueltos-de-la-indeterminacion-00\"><\/span> Exerc\u00edcios resolvidos sobre indetermina\u00e7\u00e3o 0\/0<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Abaixo preparamos v\u00e1rios exerc\u00edcios resolvidos passo a passo sobre os limites de fun\u00e7\u00f5es que fornecem indetermina\u00e7\u00f5es 0\/0. Voc\u00ea pode tentar faz\u00ea-los e depois verificar a solu\u00e7\u00e3o.<\/p>\n<p> N\u00e3o se esque\u00e7a que voc\u00ea pode nos tirar qualquer d\u00favida sobre como resolver limites nos coment\u00e1rios!<\/p>\n<h3 class=\"wp-block-heading\"> Exerc\u00edcio 1<\/h3>\n<p> Calcule o limite da seguinte fun\u00e7\u00e3o racional no ponto x=-2. <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-75edf8ebdda678fce2752f4ee280e8de_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to -2}\\frac{x^2 +2x}{x^2-x-6}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"125\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E6F9EF\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E6F9EF\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Logicamente, primeiro tentamos resolver o limite:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-9698cb3f234f704a80727c0a46642932_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to -2} \\frac{x^2 +2x}{x^2-x-6}=\\frac{(-2)^2+2\\cdot (-2)}{(-2)^2-(-2)-6}=\\frac{4-4}{4+2-6}=\\frac{0}{0}\" title=\"Rendered by QuickLaTeX.com\" height=\"44\" width=\"418\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Mas acabamos com indetermina\u00e7\u00e3o 0\/0. Devemos, portanto, fatorar os polin\u00f4mios do numerador e do denominador:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1dce911a3cd5359ea3b28e3e42159de9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to -2}\\frac{x^2 +2x}{x^2-x-6}=\\lim_{x \\to -2}\\frac{x(x+2)}{(x+2)(x-3)}\" title=\"Rendered by QuickLaTeX.com\" height=\"44\" width=\"303\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Agora simplificamos a fra\u00e7\u00e3o removendo os par\u00eanteses que se repetem no numerador e no denominador:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c38f52ab7f40e158b0a100bd9768a5d2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to -2}\\frac{x\\cancel{(x+2)}}{\\cancel{(x+2)}(x-3)}=\\lim_{x \\to -2}\\frac{x}{x-3}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"263\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E por fim, recalculamos o limite com a fra\u00e7\u00e3o simplificada: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-30649d52cdf83b256558d41b7b4ccaf5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to -2}\\frac{x}{x-3}=\\cfrac{-2}{-2-3}=\\cfrac{-2}{-5}=\\mathbf{\\cfrac{2}{5}}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"258\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Exerc\u00edcio 2<\/h3>\n<p> Resolva o limite da seguinte fun\u00e7\u00e3o quando x se aproxima de -1: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-72320d2638e9bc82cfc3f5f27b57857d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to -1}\\frac{x^3+2x^2-x-2}{x^3-5x^2+2x+8}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"182\" style=\"vertical-align: -14px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E6F9EF\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E6F9EF\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Primeiro tentamos resolver o limite normalmente:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a8e730787abc8d8843e4a818b84ee19e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to -1}\\frac{x^3+2x^2-x-2}{x^3-5x^2+2x+8}=\\frac{(-1)^3+2(-1)^2-(-1)-2}{(-1)^3-5(-1)^2+2(-1)+8} =\\frac{0}{0}\" title=\"Rendered by QuickLaTeX.com\" height=\"44\" width=\"462\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Mas obtemos a indetermina\u00e7\u00e3o 0 entre 0. Devemos, portanto, fatorar os 2 polin\u00f4mios da fra\u00e7\u00e3o:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-bab10366d036af48209c29fa26582f3e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to -1}\\frac{x^3+2x^2-x-2}{x^3-5x^2+2x+8}=\\lim_{x \\to -1}\\frac{(x-1)(x+1)(x+2)}{(x+1)(x-2)(x-4)}\" title=\"Rendered by QuickLaTeX.com\" height=\"44\" width=\"415\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Agora podemos simplificar os polin\u00f4mios:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3e66552225f948ec300edb84385118a9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to -1}\\frac{(x-1)\\cancel{(x+1)}(x+2)}{\\cancel{(x+1)}(x-2)(x-4)}=\\lim_{x \\to -1}\\frac{(x-1)(x+2)}{(x-2)(x-4)}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"386\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E resolvemos o limite: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-91295719745ca73b6874a7ebbf382bd6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to -1} \\frac{(x-1)(x+2)}{(x-2)(x-4)}=\\frac{ (-1-1)(-1+2)}{(-1-2)(-1-4)}=\\frac{(-2)\\cdot (1)}{(-3)\\cdot (-5)}=\\frac{\\bm{-2}}{\\bm{15}}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"478\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Exerc\u00edcio 3<\/h3>\n<p> Determine a solu\u00e7\u00e3o do limite da seguinte fun\u00e7\u00e3o radical: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d836a32312bb9851913a731ed3ee00e4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x\\to 2}\\frac{x^2-3x+2}{2-\\sqrt{2x}}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"123\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E6F9EF\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E6F9EF\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Primeiro, verificamos se o limite d\u00e1 algum tipo de indetermina\u00e7\u00e3o:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c9b81e0139d2863ee0c5a47c0de67ef9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x\\to 2}\\frac{x^2-3x+2}{2-\\sqrt{2x}}=\\frac{2^2-3\\cdot2+2}{2-\\sqrt{2\\cdot 2}}=\\frac{4-6+2}{2-2}=\\frac{0}{0}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"383\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> O limite d\u00e1 a indetermina\u00e7\u00e3o zero dividido por zero e temos uma raiz na fun\u00e7\u00e3o. Devemos, portanto, multiplicar o numerador e o denominador da fra\u00e7\u00e3o pelo conjugado da express\u00e3o radical:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-063e060aed092d9be29b22ed951b160e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x\\to 2}\\frac{\\left(x^2-3x+2\\right)\\cdot \\left(2+\\sqrt{2x}\\right)}{\\left(2-\\sqrt{2x}\\right)\\cdot \\left(2+\\sqrt{2x}\\right)}\" title=\"Rendered by QuickLaTeX.com\" height=\"50\" width=\"232\" style=\"vertical-align: -21px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> O denominador corresponde ao desenvolvimento da identidade not\u00e1vel do produto de uma soma e uma diferen\u00e7a, podemos portanto simplific\u00e1-lo: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eee18c8d1ddc9d4b6ccffd46123d23b0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x\\to 2}\\frac{\\left(x^2-3x+2\\right)\\cdot \\left(2+\\sqrt{2x}\\right)}{2^2-\\left(\\sqrt{2x}\\right)^2}\" title=\"Rendered by QuickLaTeX.com\" height=\"53\" width=\"232\" style=\"vertical-align: -24px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a2bf66723dd14d247966d1e5a39455fe_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x\\to 2}\\frac{\\left(x^2-3x+2\\right)\\cdot \\left(2+\\sqrt{2x}\\right)}{4-2x}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"232\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> No entanto, ainda n\u00e3o podemos simplificar os termos da fra\u00e7\u00e3o. Devemos, portanto, fatorar os polin\u00f4mios:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3fd2ce5086ce32fbee964b19d6e89b2b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x\\to 2}\\frac{\\left(x^2-3x+2\\right)\\cdot \\left(2+\\sqrt{2x}\\right)}{4-2x}=\\lim_{x\\to 2}\\frac{(x-1)(x-2)\\cdot\\left(2+\\sqrt{2x}\\right)}{-2(x-2)}\" title=\"Rendered by QuickLaTeX.com\" height=\"46\" width=\"494\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Desta forma podemos simplificar a fra\u00e7\u00e3o:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-714622fdaaae18830b983aafc10d59a3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x\\to 2}\\frac{(x-1)\\cancel{(x-2)}\\left(2+\\sqrt{2x}\\right)}{-2\\cancel{(x-2)}}=\\lim_{x\\to 2}\\frac{(x-1)\\left(2+\\sqrt{2x}\\right)}{-2}\" title=\"Rendered by QuickLaTeX.com\" height=\"46\" width=\"423\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E agora podemos determinar o resultado do limite: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0cdb8a8503f2cca27da58166341770e5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x\\to 2}\\frac{(x-1)\\left(2+\\sqrt{2x}\\right)}{-2}=\\frac{(2-1)\\left(2+\\sqrt{2\\cdot 2}\\right)}{-2}=\\frac{1\\cdot (2+2)}{-2}=\\bm{-2}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"497\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Exerc\u00edcio 4<\/h3>\n<p> Calcule o limite quando x se aproxima de 0 da seguinte fun\u00e7\u00e3o radical: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1d51d72812833040f2ae7849fde8a200_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x\\to 0}\\frac{x^2+6x}{3-\\sqrt{4x+9}}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"129\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E6F9EF\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E6F9EF\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Primeiro, tentamos calcular o limite da fun\u00e7\u00e3o como sempre fazemos:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-cc916be15d7628e8d06e6e7e1599497d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x\\to 0}\\frac{x^2+6x}{3-\\sqrt{4x+9}}=\\frac{0+0}{3-\\sqrt{4\\cdot 0+9}}=\\frac{0}{3-3}=\\frac{0}{0}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"365\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Mas obtemos a forma indeterminada de 0\/0. Portanto, multiplicamos o numerador e o denominador da fun\u00e7\u00e3o pelo conjugado da express\u00e3o irracional:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-78fa62a7006cedf8722392497434d9e4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x\\to 0}\\frac{\\left(x^2+6x\\right)\\cdot\\left(3+\\sqrt{4x+9}\\right)}{\\left(3-\\sqrt{4x+9}\\right)\\cdot\\left(3+\\sqrt{4x+9}\\right)}\" title=\"Rendered by QuickLaTeX.com\" height=\"49\" width=\"268\" style=\"vertical-align: -20px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Aplicamos a f\u00f3rmula de identidade not\u00e1vel correspondente para simplificar o denominador: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0836db36961bb85ab26e94b3a2dc9f8b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x\\to 0}\\frac{\\left(x^2+6x\\right)\\cdot\\left(3+\\sqrt{4x+9}\\right)}{3^2-\\left(\\sqrt{4x+9}\\right)^2}\" title=\"Rendered by QuickLaTeX.com\" height=\"52\" width=\"232\" style=\"vertical-align: -23px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ae003c861a15eaf19b8b7b2babd9aca1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x\\to 0}\\frac{\\left(x^2+6x\\right)\\cdot\\left(3+\\sqrt{4x+9}\\right)}{9-(4x+9)}\" title=\"Rendered by QuickLaTeX.com\" height=\"46\" width=\"232\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7fbf174fd70460599e41408dba1cf1da_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x\\to 0}\\frac{\\left(x^2+6x\\right)\\cdot\\left(3+\\sqrt{4x+9}\\right)}{-4x}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"232\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Agora fatoramos o bin\u00f4mio do numerador tomando o fator comum:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f1a55c3d856552106a7f81eb9bcc6eff_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x\\to 0}\\frac{\\left(x^2+6x\\right)\\cdot\\left(3+\\sqrt{4x+9}\\right)}{-4x}=\\lim_{x\\to 0}\\frac{\\bigl[x(x+6)\\bigr]\\cdot\\left(3+\\sqrt{4x+9}\\right)}{-4x}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"495\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Simplificamos os fatores que se repetem no numerador e no denominador da fun\u00e7\u00e3o:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-9d3996ef26d5889e65bf941fc268ed93_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x\\to 0}\\frac{\\cancel{x}\\left(x+6\\right)\\left(3+\\sqrt{4x+9}\\right)}{-4\\cancel{x}}=\\lim_{x\\to 0}\\frac{(x+6)\\left(3+\\sqrt{4x+9}\\right)}{-4}\" title=\"Rendered by QuickLaTeX.com\" height=\"40\" width=\"443\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E, finalmente, resolvemos o limite da fun\u00e7\u00e3o: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1b4874df2f48ad131d48c4e5923a5b02_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}\\displaystyle\\lim_{x\\to 0}\\frac{(x+6)\\left(3+\\sqrt{4x+9}\\right)}{-4}=\\\\[3ex]\\displaystyle=\\frac{(0+6)\\left(3+\\sqrt{4\\cdot 0+9}\\right)}{-4}=\\\\[3ex]\\displaystyle=\\frac{6\\cdot (3+3)}{-4}=\\frac{36}{-4}=\\bm{-9}\\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"153\" width=\"222\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Exerc\u00edcio 5<\/h3>\n<p> Resolva o seguinte limite usando o m\u00e9todo de indetermina\u00e7\u00e3o 0\/0:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-351bbe422f066b4d599d9c71145555ae_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to -1}\\frac{x^3+2x^2-x-2}{x^3+5x^2+7x+3}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"182\" style=\"vertical-align: -14px;\"><\/p>\n<\/p>\n<p> <span style=\"color:#ff951b\">\u27a4<\/span> <strong>Veja:<\/strong> <span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/pt\/limites-laterais\/\">como calcular os limites laterais de uma fun\u00e7\u00e3o<\/a><\/span> <\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E6F9EF\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E6F9EF\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Primeiro tentamos resolver o limite:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3ac5769f9f54b659b8472de66387df17_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to -1}\\frac{x^3+2x^2-x-2}{x^3+5x^2+7x+3}=\\frac{(-1)^3+2(-1)^2-(-1)-2}{(-1)^3+5(-1)^2+7(-1)+3}=\\frac{0}{0}\" title=\"Rendered by QuickLaTeX.com\" height=\"44\" width=\"462\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Mas no limite obtemos indetermina\u00e7\u00e3o zero sobre zero. Portanto, fatoramos os polin\u00f4mios do numerador e do denominador:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-db8f9479bc9c37f7aab1b1547eb85040_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to -1}\\frac{x^3+2x^2-x-2}{x^3+5x^2+7x+3}=\\lim_{x \\to -1}\\frac{(x-1)(x+1)(x+2)}{(x+1)^2(x+3)}\" title=\"Rendered by QuickLaTeX.com\" height=\"44\" width=\"415\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Agora simplificamos a fra\u00e7\u00e3o eliminando os fatores que se repetem no numerador e no denominador:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d338fd493df4716cc935542aa9caa99b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to -1} \\cfrac{(x-1)\\cancel{(x+1)}(x+2)}{(x+1)^{\\cancel{2}}(x+3)}=\\lim_{x \\to -1}\\cfrac{(x-1)(x+2)}{(x+1)(x+3)}\" title=\"Rendered by QuickLaTeX.com\" height=\"45\" width=\"384\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E calculamos o limite novamente:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7e67777781db8e107e153fd445404f40_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to -1}\\frac{(x-1)(x+2)}{(x+1)(x+3)}=\\frac{(-1-1)(-1+2)}{(-1+1)(-1+3)}=\\frac{-2\\cdot 1}{0 \\cdot 2}=\\frac{-2}{0} =\\infty\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"479\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Mas agora nos encontramos com a indetermina\u00e7\u00e3o de um n\u00famero dividido por 0. Devemos, portanto, calcular os limites laterais da fun\u00e7\u00e3o quando x tende a -1.<\/p>\n<p class=\"has-text-align-left\"> Primeiro resolvemos o limite lateral da fun\u00e7\u00e3o no ponto x=-1 \u00e0 esquerda:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-9d9a116a8f65f341ea8a4e497d8687d0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to -1^{-}}\\frac{(x-1)(x+2)}{(x+1)(x+3)}=\\frac{(-1-1)\\cdot (-1+2)}{(-1+1)\\cdot (-1+3)}=\\frac{-2}{-0}=+\\infty\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"444\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E ent\u00e3o calculamos o limite lateral da fun\u00e7\u00e3o no ponto x=-1 \u00e0 direita:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3ebfe6934dca4dae43c3c305708e1965_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\lim_{x \\to -1^{+}}\\frac{(x-1)(x+2)}{(x+1)(x+3)}=\\frac{(-1-1)\\cdot (-1+2)}{(-1+1)\\cdot (-1+3)}=\\frac{-2}{+0}=-\\infty\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"444\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Portanto, como os dois limites laterais n\u00e3o coincidem, o limite da fun\u00e7\u00e3o em x=-1 n\u00e3o existe: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0d26d76501f64d72e6b980c160f7858b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\displaystyle \\lim_{x \\to -1^-}f(x)= +\\infty\\neq\\lim_{x \\to -1^+}f(x)=-\\infty\\ \\bm{\\longrightarrow} \\ \\cancel{\\exists} \\lim_{x \\to -1} f(x)\" title=\"Rendered by QuickLaTeX.com\" height=\"28\" width=\"436\" style=\"vertical-align: -14px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Neste artigo explicamos como salvar o limite de uma fun\u00e7\u00e3o quando ela d\u00e1 incerteza 0\/0. Al\u00e9m disso, voc\u00ea poder\u00e1 praticar com exerc\u00edcios resolvidos sobre a indetermina\u00e7\u00e3o de zero entre zero. Como resolver a indetermina\u00e7\u00e3o zero entre zero (0\/0) Veremos ent\u00e3o como calcular o limite de uma fun\u00e7\u00e3o quando ela d\u00e1 indetermina\u00e7\u00e3o zero entre zero (0\/0). &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathority.org\/pt\/zero-entre-zero-0-0-indeterminacao\/\"> <span class=\"screen-reader-text\">Indetermina\u00e7\u00e3o zero entre zero (0\/0)<\/span> Leia mais &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"","footnotes":""},"categories":[16],"tags":[],"class_list":["post-18","post","type-post","status-publish","format-standard","hentry","category-limites-de-funcao"],"yoast_head":"<!-- This site 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