{"id":9929,"date":"2022-02-03T14:51:56","date_gmt":"2022-02-03T09:21:56","guid":{"rendered":"https:\/\/mathemerize.com\/?p=9929"},"modified":"2022-02-03T18:31:30","modified_gmt":"2022-02-03T13:01:30","slug":"factorisation-method-to-solve-limits","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/factorisation-method-to-solve-limits\/","title":{"rendered":"Factorisation Method to Solve Limits"},"content":{"rendered":"

Here you will learn what is the factorisation method to solve limits with examples.<\/p>\n

Let’s begin –<\/p>\n

Factorisation Method to Solve Limits<\/h2>\n

Consider the following limit :<\/p>\n

\\(\\displaystyle{\\lim_{x \\to a}}\\) \\(f(x)\\over g(x)\\)<\/p>\n

If by substituting x = a, \\(f(x)\\over g(x)\\), reduces to the form \\(0\\over 0\\), then (x – a) is a factor of f(x) and g(x) both.<\/p>\n

So, we first factorize f(x) and g(x) and then cancel the common factor to evaluate the limit.<\/p>\n

Also Read : How to Solve Indeterminate Forms of Limits<\/a><\/p>\n

Following algorithm may be used to evaluate the limit by factorisation method.<\/p>\n

Algorithm<\/strong> :<\/p>\n

1). Obtain the problem, say, \\(\\displaystyle{\\lim_{x \\to a}}\\) \\(f(x)\\over g(x)\\), where \\(\\displaystyle{\\lim_{x \\to a}}\\) f(x) = 0 and \\(\\displaystyle{\\lim_{x \\to a}}\\) g(x) = 0.<\/p>\n

2). Factorize f(x) and g(x).<\/p>\n

3). Cancel out the common factor.<\/p>\n

4). Use direct substitution method to obtain the limit.<\/p><\/blockquote>\n

Example<\/strong><\/span> : Evaluate : \\(\\displaystyle{\\lim_{x \\to 2}}\\) \\(x^2 – 5x + 6\\over x^2 – 4\\).<\/p>\n

Solution<\/strong><\/span> : When x = 2 the expression \\(x^2 – 5x + 6\\over x^2 – 4\\) assumes the indeterminate form \\(0\\over 0\\).<\/p>\n

Therefore, (x – 2) is a common factor in numerator and denominator.<\/p>\n

Factorising the numerator and denominator, we have<\/p>\n

\\(\\displaystyle{\\lim_{x \\to 2}}\\) \\(x^2 – 5x + 6\\over x^2 – 4\\) = \\(\\displaystyle{\\lim_{x \\to 2}}\\) \\((x – 2)(x – 3)\\over (x + 2)(x – 2)\\)<\/p>\n

= \\(\\displaystyle{\\lim_{x \\to 2}}\\) \\(x – 3\\over x + 2\\) = \\(2 – 3\\over 2 + 2\\) = \\(-1\\over 4\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

Here you will learn what is the factorisation method to solve limits with examples. Let’s begin – Factorisation Method to Solve Limits Consider the following limit : \\(\\displaystyle{\\lim_{x \\to a}}\\) \\(f(x)\\over g(x)\\) If by substituting x = a, \\(f(x)\\over g(x)\\), reduces to the form \\(0\\over 0\\), then (x – a) is a factor of f(x) …<\/p>\n

Factorisation Method to Solve Limits<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","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":""},"categories":[19],"tags":[],"yoast_head":"\nFactorisation Method to Solve Limits - Mathemerize<\/title>\n<meta name=\"description\" content=\"In this method you will learn what is the factorisation method to solve limits with examples.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/factorisation-method-to-solve-limits\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Factorisation Method to Solve Limits - 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