Evaluating the limit $\lim_{x\to\infty}\left(\frac {x^5+\pi x^4+e}{x^5+ex^4+\pi}\right)^x$ How can I evaluate the following limit?
$$\lim_{x\to\infty}\left(\frac {x^5+\pi x^4+e}{x^5+ex^4+\pi}\right)^x$$
To solve this limit, I divided the numerator and denominator by $x^5$ and I got
$$\lim_{x\to\infty}\left(\frac {1+\frac {1}{x}\pi+\frac {1}{x^5}e}{1+\frac {1}{x}e+\frac {1}{x^5}\pi}\right)^x$$
But I realized when $x\to\infty$  the expression is still indeterminate.
I tried changing the $x=\frac {1}{y}$ variable to be able to apply Lophital's rule, but I couldn't get a useful result.
 A: The limit is $e^{\pi-e},$ because $A^x=e^{x\ln A}$ ($\forall A>0$) and as $x\to\infty,$
$$\begin{align}x\ln\frac {1+\frac\pi x+o(1/x)}{1+\frac ex+o(1/x)}&=x\Big(\ln(1+\pi/x+o(1/x))-\ln(1+e/x+o(1/x))\Big)\\&=x\Big((\pi/x+o(1/x))-(e/x+o(1/x))\Big)\\&=x\left(\frac{\pi-e}x+o(1/x)\right)\\&\to\pi-e.
\end{align}$$
A: So this is the typical $1^\infty$ type-limit. To find the limit I always start by adding and subtracting $1$ inside the parenthesis (you are "apparentlly" doing nothing), like so $$\lim_{x\rightarrow \infty}\Bigg(1-1+\frac{x^5+\pi x^4+e}{x^5+e x^4+\pi}\Bigg)^x=\lim_{x\rightarrow \infty}\Bigg(1+\frac{(\pi-e)(x^4-1)}{x^5+e x^4+\pi}\Bigg)^x=(\star)$$
Since we know that $\lim_{x\rightarrow \infty}(1+\frac{1}{x})^x=e$, our priority is to set our limit in a similar way.
$$(\star)=\lim_{x\rightarrow \infty}\Bigg(1+\frac{1}{\frac{x^5+e x^4+\pi}{(\pi-e)(x^4-1)}}\Bigg)^x=\lim_{x\rightarrow \infty}\Bigg(\Bigg[1+\frac{1}{\frac{x^5+e x^4+\pi}{(\pi-e)(x^4-1)}}\Bigg]^{\frac{x^5+e x^4+\pi}{(\pi-e)(x^4-1)}}\Bigg)^{x\cdot\frac{(\pi-e)(x^4-1)}{x^5+e x^4+\pi}}$$
Please go ahead and see by yourself that this last limit is equal to the first one you gave. If not, trust me. Now, we know that the part inside the parenthesis converges to $e$ $$\lim_{x\rightarrow \infty}\Bigg[1+\frac{1}{\frac{x^5+e x^4+\pi}{(\pi-e)(x^4-1)}}\Bigg]^{\frac{x^5+e x^4+\pi}{(\pi-e)(x^4-1)}}=e$$ and our exponent converges to $\pi-e$ since $$\lim_{x\rightarrow \infty}x\cdot\frac{(e-\pi)(x^4-1)}{x^5+e x^4+\pi}=\pi-e$$
And so the limit is $e^{\pi-e}$. If you found this answer helpful please flag it as a correct answer. Thanks!!
A: $$\lim_{x\to\infty}\left(\frac {1+\frac {1}{x}\pi+\frac {1}{x^5}e}{1+\frac {1}{x}e+\frac {1}{x^5}\pi}\right)^x$$
Arrived at this point, you can use geometric series:
$$\left(\frac {1+\frac {1}{x}\pi+\frac {1}{x^5}e}{1+\frac {1}{x}e+\frac {1}{x^5}\pi}\right)^x\sim\left(\left(1+\frac {1}{x}\pi+\frac {1}{x^5}e\right)\left(1-\frac {1}{x}e-\frac {1}{x^5}\pi \right)\right)^x\sim \left(1 +\frac{\pi-e}{x}\right)^x \to e^{\pi-e}$$
