Simple Prooof $\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{1}{x} \right)}^{x}}=e\,\,\,\,,\,\,\forall x\in \mathbb{R}$

Exercise:

Show that , $\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{1}{x} \right)}^{x}}=e\,\,\,\,,\,\,\forall x\in \mathbb{R}$

Proof: Let $y=\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{1}{x} \right)}^{x}}\Leftrightarrow \ln y=\underset{x\to \infty }{\mathop{\lim }}\,x\ln \left( 1+\frac{1}{x} \right)=0\times \infty $ind form

So $\ln y=\underset{x\to \infty }{\mathop{\lim }}\,x\ln \left( 1+\frac{1}{x} \right)=\underset{x\to \infty }{\mathop{\lim }}\,\frac{\ln \left( 1+\frac{1}{x} \right)}{\frac{1}{x}}$

Take $u=\frac{1}{x}$ as $x\to \infty \,,u\to 0$

So $\underset{x\to \infty }{\mathop{\lim }}\,\frac{\ln \left( 1+\frac{1}{x} \right)}{\frac{1}{x}}=\underset{u\to 0}{\mathop{\lim }}\,\frac{\ln \left( 1+u \right)}{u}=\frac{0}{0}\,$ind form

Take $h=u+1$ as $u\to 0,h\to 1$

So $\underset{u\to 0}{\mathop{\lim }}\,\frac{\ln \left( 1+u \right)}{u}=\underset{h\to 1}{\mathop{\lim }}\,\frac{\ln \left( h \right)}{h-1}=\underset{h\to 1}{\mathop{\lim }}\,\frac{\ln \left( h \right)-\ln \left( 1 \right)}{h-1}=\frac{d}{dh}{{\left( \ln \left( h \right) \right)}_{h=1}}=1$

Hence $\ln y=1\Leftrightarrow y=e\Leftrightarrow \underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{1}{x} \right)}^{x}}=e,\,\,\forall x\in \mathbb{R}$