# Limits, Continuity and Differentiability – Method of Rationalization

## Limits, Continuity and Differentiability – Method of Rationalization

Rationalization method is used when, we have radical signs in an expression (like ½, ⅓ etc) and there exists a negative sign between two terms of an algebraic expression.

After rationalization, the terms are factorized which on cancellation gives the required result.

Example: $$\underset{x\to \,\pm \,\infty }{\mathop{\lim }}\,x\left( \sqrt{{{x}^{2}}+k}-x \right),\,k>0$$ is equal to

Solution: $$\underset{x\to \,\pm \,\infty }{\mathop{\lim }}\,x\left( \sqrt{{{x}^{2}}+k}-x \right)\frac{\left( \sqrt{{{x}^{2}}+k}+x \right)}{\sqrt{{{x}^{2}}+k}+x}$$

$$=\underset{x\to \,\pm \,\infty }{\mathop{\lim }}\,\frac{x\left( {{x}^{2}}+k-{{x}^{2}} \right)}{\left( \sqrt{{{x}^{2}}+k}+x \right)}=\underset{x\to \,\pm \,\infty }{\mathop{\lim }}\,\frac{xk}{\left( |x|\sqrt{\left( 1+\frac{k}{{{x}^{2}}} \right)} \right)+x}$$

Here, we have to consider two cases.

i) When x → ∞; |x| = x

Then, the given limit $$=\underset{x\to \infty }{\mathop{\lim }}\,\frac{xk}{x\sqrt{\left( 1+\frac{k}{{{x}^{2}}} \right)}+x}$$

$$=\underset{x\to \infty }{\mathop{\lim }}\,\frac{xk}{x\left( \sqrt{\left( 1+\frac{k}{{{x}^{2}}} \right)}+1 \right)}=\frac{k}{2}$$

ii) When x → -∞; |x| = -x

Then, we have$$\underset{x\to -\infty }{\mathop{\lim }}\,\frac{xk}{-x\sqrt{\left( 1+\frac{k}{{{x}^{2}}} \right)}+x}$$

$$\underset{x\to -\infty }{\mathop{\lim }}\,\frac{xk}{x\left( -\sqrt{\left( 1+\frac{k}{{{x}^{2}}} \right)}+1 \right)}=\frac{k}{-{{1}^{-}}+1}=\frac{k}{{{0}^{-}}}\to -\infty$$.