# 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

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

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