{"id":62,"date":"2023-09-17T11:09:13","date_gmt":"2023-09-17T11:09:13","guid":{"rendered":"https:\/\/mathority.org\/nl\/nul-tussen-nul-0-0-onbepaaldheid-2\/"},"modified":"2023-09-17T11:09:13","modified_gmt":"2023-09-17T11:09:13","slug":"nul-tussen-nul-0-0-onbepaaldheid-2","status":"publish","type":"post","link":"https:\/\/mathority.org\/nl\/nul-tussen-nul-0-0-onbepaaldheid-2\/","title":{"rendered":"Nul onbepaaldheid tussen nul (0\/0)"},"content":{"rendered":"<p>In dit artikel leggen we uit hoe je de limiet van een functie kunt opslaan als deze de onzekerheid 0\/0 geeft. Daarnaast kun je oefenen met opgeloste oefeningen op de onbepaaldheid van nul tussen nul. <\/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> Hoe nul-onbepaaldheid tussen nul (0\/0) op te lossen<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> We zullen dan zien hoe we de limiet van een functie kunnen berekenen wanneer deze een onbepaaldheid nul geeft tussen nul (0\/0). Om dit te doen, zullen we stap voor stap een voorbeeld berekenen:<\/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> We proberen eerst de limiet te berekenen door de waarde van x in de functie te vervangen:<\/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> Maar we verkrijgen de onbepaaldheid 0 gedeeld door 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\"> Wanneer de limiet van een puntfunctie de <strong>onzekerheid 0\/0<\/strong> geeft, is het noodzakelijk om de polynomen van de teller en de noemer in factoren te ontbinden en vervolgens de gemeenschappelijke factoren te vereenvoudigen.<\/p>\n<\/div>\n<p> We moeten daarom de polynomen van de teller en de noemer van de breuk ontbinden in factoren. Om dit te doen, gebruiken we de regel van 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=\"onbepaaldheidsfactorisatie 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> Als je niet weet <u style=\"text-decoration-color:#FF9B28;\">hoe je een polynoom moet ontbinden<\/u> , raden we je aan de uitleg te bekijken op onze site gespecialiseerd in polynomen: <u style=\"text-decoration-color:#FF9B28;\">www.polinomios.org<\/u><\/p>\n<p> Dus zodra de polynomen in factoren zijn verwerkt, is de limiet als volgt:<\/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> We kunnen nu de limiet vereenvoudigen door de factoren te elimineren die zich herhalen in de teller en de noemer van de breuk:<\/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\"> En ten slotte herberekenen we de limiet:<\/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> Zoals je kunt zien, is het, zodra we de polynomen ontbinden en vereenvoudigen, heel gemakkelijk om de oplossing in de limiet te vinden. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"indeterminacion-00-con-raices\"><\/span>Onbepaaldheid 0\/0 met wortels<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> We hebben zojuist gezien hoe de 0\/0 onbepaaldheid van rationale functies wordt opgelost. Als de limiet echter een irrationele (of radicale) functie heeft, wordt de 0\/0-onbepaaldheid anders opgelost.<\/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> Eerst proberen we de limiet op te lossen door de volgende bewerkingen uit te voeren:<\/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> Maar we krijgen nul boven nul onbepaaldheid. <\/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\"> Als de <strong>limiet van een functie met wortels de onbepaaldheid 0\/0 oplevert<\/strong> , moet je de teller en de noemer van de breuk vermenigvuldigen met de conjugaat van de worteluitdrukking.<\/p>\n<\/div>\n<p> \u27a4 Onthoud dat de conjugaat dezelfde irrationele uitdrukking is, maar met een aangepast middelste teken.<\/p>\n<p> Vervolgens vermenigvuldigen we zowel de teller als de noemer van de breuk met het conjugaat van de worteluitdrukking:<\/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> Binnen dit soort grenzen zullen we door deze stap te doen altijd een opmerkelijke identiteit verkrijgen die we kunnen vereenvoudigen. In dit geval hebben we in de noemer het product van een som en een verschil, dus:<\/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> We vereenvoudigen de factor die wordt herhaald in de teller en de noemer:<\/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> En op deze manier kunnen we het resultaat van de limiet vinden: <\/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> Opgeloste oefeningen over onbepaaldheid 0\/0<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Hieronder hebben we verschillende stapsgewijze opgeloste oefeningen voorbereid over de limieten van functies die 0\/0 onbepaaldheid opleveren. U kunt proberen ze uit te voeren en vervolgens de oplossing controleren.<\/p>\n<p> Vergeet niet dat je ons in de reacties al je vragen over het oplossen van limieten kunt stellen!<\/p>\n<h3 class=\"wp-block-heading\"> Oefening 1<\/h3>\n<p> Bereken de limiet van de volgende rationale functie op het punt 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>Zie de oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Logischerwijs proberen we eerst de limiet op te lossen:<\/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\"> Maar we eindigen met 0\/0 onbepaaldheid. We moeten daarom de polynomen van de teller en de noemer ontbinden in factoren:<\/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\"> Nu vereenvoudigen we de breuk door de haakjes te verwijderen die herhaald worden in de teller en de noemer:<\/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\"> En ten slotte herberekenen we de limiet met de vereenvoudigde breuk: <\/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\">Oefening 2<\/h3>\n<p> Los de limiet van de volgende functie op als x -1 nadert: <\/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>Zie de oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> We proberen eerst de limiet op te lossen zoals gewoonlijk:<\/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\"> Maar we verkrijgen de onbepaaldheid 0 tussen 0. We moeten daarom de 2 polynomen van de breuk ontbinden:<\/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\"> We kunnen nu de polynomen vereenvoudigen:<\/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\"> En we lossen de limiet op: <\/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\">Oefening 3<\/h3>\n<p> Bepaal de oplossing van de limiet van de volgende radicale functie: <\/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>Zie de oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Eerst controleren we of de limiet enige vorm van onbepaaldheid geeft:<\/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\"> De limiet geeft de onbepaaldheid nul gedeeld door nul en we hebben een wortel in de functie. We moeten daarom de teller en de noemer van de breuk vermenigvuldigen met de conjugaat van de worteluitdrukking:<\/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\"> De noemer komt overeen met de ontwikkeling van de opmerkelijke identiteit van het product van een som en een verschil, we kunnen het daarom vereenvoudigen: <\/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\"> We kunnen de voorwaarden van de breuk echter nog niet vereenvoudigen. We moeten daarom de polynomen in factoren ontbinden:<\/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\"> Op deze manier kunnen we de breuk vereenvoudigen:<\/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\"> En nu kunnen we het resultaat van de limiet bepalen: <\/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\">Oefening 4<\/h3>\n<p> Bereken de limiet als x 0 nadert van de volgende radicale functie: <\/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>Zie de oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Eerst proberen we de limiet van de functie te berekenen zoals we altijd doen:<\/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\"> Maar we krijgen de onbepaalde vorm van 0\/0. Daarom vermenigvuldigen we de teller en de noemer van de functie met de conjugaat van de irrationele uitdrukking:<\/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\"> We passen de overeenkomstige opmerkelijke identiteitsformule toe om de noemer te vereenvoudigen: <\/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\"> Nu ontbinden we de binomiaal van de teller door de gemeenschappelijke factor te nemen:<\/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\"> We vereenvoudigen de factoren die worden herhaald in de teller en de noemer van de functie:<\/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\"> En ten slotte lossen we de limiet van de functie op: <\/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\">Oefening 5<\/h3>\n<p> Los de volgende limiet op met behulp van de 0\/0 onbepaaldheidsmethode:<\/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>Zie:<\/strong> <span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/nl\/laterale-grenzen\/\">hoe bereken je de laterale grenzen van een functie<\/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>Zie de oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> We proberen eerst de limiet op te lossen:<\/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\"> Maar in de limiet verkrijgen we nul-op-nul onbepaaldheid. Daarom ontbinden we de polynomen van de teller en de noemer in factoren:<\/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\"> Nu vereenvoudigen we de breuk door de factoren te elimineren die zich herhalen in de teller en de noemer:<\/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\"> En we berekenen de limiet opnieuw:<\/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\"> Maar nu bevinden we ons met de onbepaaldheid van een getal gedeeld door 0. We moeten daarom de laterale grenzen van de functie berekenen wanneer x naar -1 neigt.<\/p>\n<p class=\"has-text-align-left\"> We lossen eerst de laterale limiet van de functie op in het punt x=-1 aan de linkerkant:<\/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\"> En dan berekenen we de laterale limiet van de functie op het punt x=-1 aan de rechterkant:<\/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\"> Omdat de twee laterale limieten niet samenvallen, bestaat de limiet van de functie op x = -1 niet: <\/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>In dit artikel leggen we uit hoe je de limiet van een functie kunt opslaan als deze de onzekerheid 0\/0 geeft. Daarnaast kun je oefenen met opgeloste oefeningen op de onbepaaldheid van nul tussen nul. Hoe nul-onbepaaldheid tussen nul (0\/0) op te lossen We zullen dan zien hoe we de limiet van een functie kunnen &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathority.org\/nl\/nul-tussen-nul-0-0-onbepaaldheid-2\/\"> <span class=\"screen-reader-text\">Nul onbepaaldheid tussen nul (0\/0)<\/span> Lees meer &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":[43],"tags":[],"class_list":["post-62","post","type-post","status-publish","format-standard","hentry","category-functielimieten"],"yoast_head":"<!-- This site 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