Limit exercise asked in the math group and solved without using L'Hospital Rule

Exercise:

Find $\underset{x\to 0}{\mathop{\lim }}\,{{\left( \tan \left( x+\frac{\pi }{4} \right) \right)}^{1/x}}$

Solution: Let $y={{\left( \tan \left( x+\frac{\pi }{4} \right) \right)}^{1/x}}\Leftrightarrow \ln y=\frac{1}{x}\ln \left( \tan \left( x+\frac{\pi }{4} \right) \right)$

So \(\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{x}\ln \left( \tan \left( x+\frac{\pi }{4} \right) \right)=\frac{0}{0}\,\,ind\,form\)

Observe that $\underset{x\to 0}{\mathop{\lim }}\,\frac{\ln \left( \tan \left( x+\frac{\pi }{4} \right) \right)}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\ln \left( \tan \left( x+\frac{\pi }{4} \right) \right)-\ln \left( \tan \frac{\pi }{4} \right)}{x-0}$

$=\underset{x\to 0}{\mathop{\lim }}\,\frac{\ln \left( \tan \left( x+\frac{\pi }{4} \right) \right)-0}{x-0}=\underset{x\to 0}{\mathop{\lim }}\,\frac{f\left( x \right)-f\left( 0 \right)}{x-0}=f'\left( 0 \right)$ where $f\left( x \right)=\ln \left( \tan \left( x+\frac{\pi }{4} \right) \right)$

But $f'\left( x \right)=\frac{{{\sec }^{2}}\left( x+\frac{\pi }{4} \right)}{\tan \left( x+\frac{\pi }{4} \right)}\Leftrightarrow f'\left( 0 \right)=\frac{{{\sec }^{2}}\left( \frac{\pi }{4} \right)}{\tan \left( \frac{\pi }{4} \right)}=\frac{2}{1}=2$

Thus $\underset{x\to 0}{\mathop{\lim }}\,y=\underset{x\to 0}{\mathop{\lim }}\,{{e}^{\frac{\ln \left( \tan \left( x+\frac{\pi }{4} \right) \right)}{x}}}={{e}^{2}}\Leftrightarrow y=\underset{x\to 0}{\mathop{\lim }}\,{{\left( \tan \left( x+\frac{\pi }{4} \right) \right)}^{1/x}}={{e}^{2}}$

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The idea of definition of differential coefficient credit to  Kunny Shogun