Limit exercise asked by Hamza Mahmood in their facebook wall

Exercise:

Show that, $\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \frac{\pi +x}{e+x} \right)}^{{{\pi }^{y}}+xy+{{y}^{e}}}}={{e}^{y\left( \pi -e \right)}}$

Solution: Let $w={{\left( \frac{\pi +x}{e+x} \right)}^{xy+{{\pi }^{y}}+{{y}^{e}}}}\Leftrightarrow \ln w=\left( xy+{{\pi }^{y}}+{{y}^{e}} \right)\ln \left( \frac{\pi +x}{e+x} \right)$

$\Rightarrow \large{{{e}^{\ln w}}={{e}^{\left( xy+{{\pi }^{y}}+{{y}^{e}} \right)\ln \left( \frac{\pi +x}{x+e} \right)}}\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,{{e}^{\ln w}}=\underset{x\to \infty }{\mathop{\lim }}\,{{e}^{\left( xy+{{\pi }^{y}}+{{y}^{e}} \right)\ln \left( \frac{\pi +x}{x+e} \right)}}}$

So $\large{w={{e}^{\underset{x\to \infty }{\mathop{\lim }}\,\left( xy+{{\pi }^{y}}+{{y}^{e}} \right)\ln \left( \frac{x+\pi }{x+e} \right)}}}$

But $\underset{x\to \infty }{\mathop{\lim }}\,\frac{\ln \left( \frac{x+\pi }{x+e} \right)}{\frac{1}{xy+{{\pi }^{y}}+{{y}^{e}}}}=\frac{0}{0}\,\,ind\,form$ so by the L’Hospital rule we get

So $\underset{x\to \infty }{\mathop{\lim }}\,\frac{\frac{e-\pi }{\left( e+x \right)\left( \pi +x \right)}}{-\frac{y}{{{\left( xy+{{\pi }^{y}}+{{y}^{e}} \right)}^{2}}}}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{-\left( e-\pi  \right){{\left( xy+{{\pi }^{y}}+{{y}^{e}} \right)}^{2}}}{y\left( e+x \right)\left( \pi +x \right)}=\frac{\pi -e}{y}\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{\left( xy+{{\pi }^{y}}+{{y}^{e}} \right)}^{2}}}{\left( e+x \right)\left( \pi +x \right)}$

But $\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{\left( xy+{{\pi }^{y}}+{{y}^{e}} \right)}^{2}}}{\left( e+x \right)\left( \pi +x \right)}=\frac{\infty }{\infty }\,\,ind\,form$ also using L’Hopital Rule,

$=\underset{x\to \infty }{\mathop{\lim }}\,\frac{2y\left( xy+{{\pi }^{y}}+{{y}^{e}} \right)}{e+\pi +2x}=2y\underset{x\to \infty }{\mathop{\lim }}\,\frac{xy+{{\pi }^{y}}+{{y}^{e}}}{e+\pi +2x}\overset{H.R}{\mathop{=}}\,2y\underset{x\to \infty }{\mathop{\lim }}\,\frac{y}{2}={{y}^{2}}$

So $\underset{x\to \infty }{\mathop{\lim }}\,\left( xy+{{\pi }^{y}}+{{y}^{e}} \right)\ln \left( \frac{x+\pi }{x+e} \right)=\left( \frac{\pi -e}{y} \right)\left( {{y}^{2}} \right)=y\left( \pi -e \right)$

Therefore $\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \frac{\pi +x}{e+x} \right)}^{{{\pi }^{y}}+xy+{{y}^{e}}}}={{e}^{y\left( \pi -e \right)}}$        Q.E.D