Limit of function

[xaranton]

Find the limit \[\displaystyle\mathop{\lim}\limits_{x\rightarrow+\infty}{\left({\frac{x^2+5x+4}{x^2-3x+7}}\right)^{x}}\,.\]

[Grigorios Kostakos]

\begin{align*}
\displaystyle\mathop{\lim}\limits_{x\rightarrow+\infty}{\left({\frac{x^2+5x+4}{x^2-3x+7}}\right)^{x}}&=\mathop{\lim}\limits_{x\rightarrow+\infty}{\exp\left({x\,\log\left({\tfrac{x^2+5x+4}{x^2-3x+7}}\right)}\right)}\\
&=\exp\left({\mathop{\lim}\limits_{x\rightarrow+\infty}{x\,\log\left({\tfrac{x^2+5x+4}{x^2-3x+7}}\right)}}\right)\\
&=\exp\left({\mathop{\lim}\limits_{x\rightarrow+\infty}{\frac{\log\left({\frac{x^2+5x+4}{x^2-3x+7}}\right)}{\frac{1}{x}}}}\right)\\
&\stackrel{\frac{0}{0}}{=}\exp\left({\mathop{\lim}\limits_{x\rightarrow+\infty}{\frac{-{\frac {8x^2-6x-47}{ \left(x^2+5x+4 \right) \left( x^2-3x+7 \right) }}
}{-\frac{1}{x^2}}}}\right)\\
&=\exp\left({\mathop{\lim}\limits_{x\rightarrow+\infty}{\frac {x^2\,(8x^2-6x-47)}{ \left(x^2+5x+4 \right) \left( x^2-3x+7 \right) }}}\right)\\
&=\exp\left({\mathop{\lim}\limits_{x\rightarrow+\infty}{\frac {8-\frac{6}{x}-\frac{47}{x^2}}{1+\frac{2}{x}-\frac{4}{x^2}+\frac{23}{x^3}+\frac{28}{x^4}}}}\right)\\
&=\exp\left(\tfrac{8}{1}\right)=e^8\,.
\end{align*}