# Limits – Part2

## Limits of the Form $$\underset{x\to a}{\mathop{\lim }}\,{{\left( f(x) \right)}^{g(x)}}$$–$${{1}^{\infty }}$$

1. $$\underset{x\to 0}{\mathop{\lim }}\,{{\left( 1+x \right)}^{\frac{1}{x}}}=e\ \ (or)\ \ \ \ \ \underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{1}{x} \right)}^{x}}=e$$

$$=\underset{x\to 0}{\mathop{\lim }}\,\left( 1+\frac{1}{x}x+\frac{\frac{1}{x}\left( \frac{1}{x}-1 \right)}{2!}{{x}^{2}}+\frac{\frac{1}{x}\left( \frac{1}{x}-1 \right)\left( \frac{1}{x}-2 \right)}{2!}{{x}^{3}}+…. \right)$$

$$=\underset{x\to 0}{\mathop{\lim }}\,\left( 1+1+\frac{1(1-x)}{2!}+\frac{1(1-x)(1-2x)}{3!}+… \right)$$,

$$=\left( 1+1+\frac{1}{2!}+\frac{1}{3!}+…. \right)=e$$

2. $$L=\underset{x\to a}{\mathop{\lim }}\,f{{\left( x \right)}^{g(x)}}$$. If $$\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=1\ \ and\ \ \underset{x\to a}{\mathop{\lim }}\,g\left( x \right)=\infty$$

Then  $$L=\underset{x\to a}{\mathop{\lim }}\,f{{\left( x \right)}^{g(x)}}$$,

$$L=\underset{x\to a}{\mathop{\lim }}\,{{\left\{ 1+\left( f\left( x \right)-1 \right) \right\}}^{\frac{1}{f\left( x \right)-1}\left( f\left( x \right)-1 \right)\times g\left( x \right)}}$$,

$$L=\underset{x\to a}{\mathop{\lim }}\,{{\left\{ {{\left( 1+\left( f\left( x \right)-1 \right) \right)}^{\frac{1}{f\left( x \right)-1}}} \right\}}^{\underset{x\to a}{\mathop{\lim }}\,}}^{\left( f\left( x \right)-1 \right)\times g\left( x \right)}$$,

$$={{e}^{\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)-1 \right)\times g\left( x \right)}}$$.

Example: Evaluate  $$\underset{x\to 0}{\mathop{\lim }}\,{{\left( 1+x \right)}^{\cos ecx}}$$

Solution:

$$\underset{x\to 0}{\mathop{\lim }}\,{{\left( 1+x \right)}^{\cos ecx}}$$,

$$\underset{x\to 0}{\mathop{\lim }}\,{{\left( 1+x \right)}^{\frac{1}{\sin x}}}$$,

$$\underset{x\to 0}{\mathop{\lim }}\,{{\left( 1+x \right)}^{\frac{1}{\sin x}\times \frac{x}{x}}}$$,

$$\underset{x\to 0}{\mathop{\lim }}\,{{\left\{ \left( 1+x \right) \right\}}^{\frac{x}{\sin x}\times \frac{1}{x}}}$$,

$$\underset{x\to 0}{\mathop{\lim }}\,{{\left\{ {{\left( 1+x \right)}^{\frac{1}{x}}} \right\}}^{\frac{x}{\sin x}}}$$,

$$\underset{x\to 0}{\mathop{\lim }}\,{{\left\{ {{\left( 1+x \right)}^{\frac{1}{x}}} \right\}}^{\underset{x\to 0}{\mathop{\lim }}\,\ \ \frac{x}{\sin x}}}$$,

$$\because \underset{x\to 0}{\mathop{\lim }}\,{{\left( 1+x \right)}^{\frac{1}{x}}}=e$$,

$${{e}^{\underset{x\to 0}{\mathop{\lim }}\,\ \ \frac{x}{\sin x}}}$$,

$$={{e}^{1}}$$.