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

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

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

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