Rationalisation Method to Solve Limits

Here you will learn what is the rationalisation method to solve or find limits with examples.

Let’s begin –

Rationalisation Method to Solve Limits

This method is particularly used when either the numerator or denominator or both involve expression consisting of square roots and substituting the value of x the rational expression takes the form $0\over 0$, $\infty\over \infty$.

Also Read : How to Solve Indeterminate Forms of Limits

Following examples illustrate the above method :

Example : Evaluate : $\displaystyle{\lim_{x \to 0}}$ $\sqrt{2 + x} – \sqrt{2}\over x$.

Solution : When x = 0, the expression $\sqrt{2 + x} – \sqrt{2}\over x$ takes the form $0\over 0$.

Rationalising the numerator we have,

$\displaystyle{\lim_{x \to 0}}$ $(\sqrt{2 + x} – \sqrt{2})(\sqrt{2 + x} + \sqrt{2})\over x(\sqrt{2 + x} + \sqrt{2})$

= $\displaystyle{\lim_{x \to 0}}$ $2 + x – 2\over x(\sqrt{2 + x} + \sqrt{2})$

= $\displaystyle{\lim_{x \to 0}}$ $1\over x(\sqrt{2 + x} + \sqrt{2})$ = $1\over 2\sqrt{2}$

Example : Evaluate the limit : $\displaystyle{\lim_{x \to 1}}$ [${4 – \sqrt{15x + 1}}\over {2 – \sqrt{3x + 1}}$]

Solution : $\displaystyle{\lim_{x \to 1}}$ [${4 – \sqrt{15x + 1}}\over {2 – \sqrt{3x + 1}}$]

Rationalising the numerator and denominator both we have,

= $\displaystyle{\lim_{x \to 1}}$ ${(4 – \sqrt{15x + 1})(2 + \sqrt{3x + 1})(4 + \sqrt{15x + 1})}\over {(2 – \sqrt{3x + 1})(4 + \sqrt{15x + 1})(2 + \sqrt{3x + 1})}$

= $\displaystyle{\lim_{x \to 1}}$ $(15 – 15x)\over {3 – 3x}$$\times$$2 + \sqrt{3x + 1}\over {4 + \sqrt{15x + 1}}$

= $5\over 2$