# 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,

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

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

= $$5\over 2$$