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

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