<s>[INST] How to solve $\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n$ Question:
$$\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n=?$$
My work:
$\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n=\lim_{n\to\infty}\left(\dfrac{n^2(1+5/n+3/n^2}{n^2(1+1/n+2/n^2)}\right)^n=\lim_{n\to\infty}\left(\dfrac{1+5/n+3/n^2}{1+1/n+2/n^2}\right)^n$
$\log L=n\log\left(\dfrac{1+5/n+3/n^2}{1+1/n+2/n^2}\right)$
Is this equal to 0? Then the answer would be $e^0=1$.

The answer was given as $e^4$ and I have no idea how to get to that.

 [/INST] Another way: Note it is easy to show for large enough $n$, (in this case $n>7$):
$$\frac4{n+1} < \frac{4n+1}{n^2+n+2} < \frac4{n}$$
$$\implies \left(1 + \frac4{n+1}\right)^n < \left(\frac{n^2+5n+3}{n^2+n+2} \right)^n < \left(1 + \frac4{n}\right)^n$$
Now squeezing does the rest.
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