Factorisation Method to Solve Limits

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) and g(x) both.

So, we first factorize f(x) and g(x) and then cancel the common factor to evaluate the limit.

Also Read : How to Solve Indeterminate Forms of Limits

Following algorithm may be used to evaluate the limit by factorisation method.

Algorithm :

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.

2). Factorize f(x) and g(x).

3). Cancel out the common factor.

4). Use direct substitution method to obtain the limit.

Example : Evaluate : $\displaystyle{\lim_{x \to 2}}$ $x^2 – 5x + 6\over x^2 – 4$.

Solution : When x = 2 the expression $x^2 – 5x + 6\over x^2 – 4$ assumes the indeterminate form $0\over 0$.

Therefore, (x – 2) is a common factor in numerator and denominator.

Factorising the numerator and denominator, we have

$\displaystyle{\lim_{x \to 2}}$ $x^2 – 5x + 6\over x^2 – 4$ = $\displaystyle{\lim_{x \to 2}}$ $(x – 2)(x – 3)\over (x + 2)(x – 2)$

= $\displaystyle{\lim_{x \to 2}}$ $x – 3\over x + 2$ = $2 – 3\over 2 + 2$ = $-1\over 4$