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,

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

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

= \(5\over 2\)

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