# Limits, Continuity and Differentiability – Method of Substitution

## Limits, Continuity and Differentiability – Method of Substitution

Method of substitution: In order to evaluate $$\underset{x\to a}{\mathop{\lim }}\,f(x)$$ we many substitute x = a + h (or) a – h, so that as x – a, h → 0.

Thus $$\underset{x\to a}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(a+h)$$ (or) $$\underset{x\to a}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(a-h)$$.

This method is applied to bring the limit at zero as the most of formulate are given as $$\underset{x\to a}{\mathop{\lim }}\,f(x)$$.

Example: Evaluate $$\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,\frac{\sqrt{\pi }-\sqrt{{{\cos }^{-1}}x}}{\sqrt{x+1}}$$.

Solution: Given that

$$\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,\frac{\sqrt{\pi }-\sqrt{{{\cos }^{-1}}x}}{\sqrt{x+1}}$$.

Put cos⁻¹ x = θ ⇒ x = cosθ

Above limits

$$\underset{\theta \to \pi }{\mathop{\lim }}\,\frac{\sqrt{\pi }-\sqrt{\theta }}{\sqrt{1+\cos \theta }}$$.

$$\underset{\theta \to \pi }{\mathop{\lim }}\,\frac{\sqrt{\pi }-\sqrt{\theta }}{\sqrt{1+\cos \theta }}\frac{\sqrt{\pi }+\sqrt{\theta }}{\sqrt{\pi }+\sqrt{\theta }}$$.

$$\underset{\theta \to \pi }{\mathop{\lim }}\,\frac{1}{\sqrt{\pi }+\sqrt{\theta }}.\underset{\theta \to \pi }{\mathop{\lim }}\,\frac{\pi -\theta }{\sqrt{2}.\cos (\theta /2)}$$.

$$\underset{k\to 0}{\mathop{\lim }}\,\frac{k}{\sqrt{2{{\cos }^{2}}\left( \frac{\pi }{2}-\frac{k}{2} \right)}}.\frac{1}{2\sqrt{\pi }}$$.

Putting θ = π – k

$$\underset{k\to 0}{\mathop{\lim }}\,\frac{k/2}{\frac{\sqrt{2}}{2}.\sin \frac{k}{2}}.\frac{1}{2\sqrt{\pi }}=\frac{1}{2\sqrt{\pi }}$$.