<s>[INST] Calculating $\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$ Calculate $$\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$$
Here is my attempt:
$$\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}= \left(\frac{4\infty^2+5\infty-6}{4\infty^2+3\infty-10}\right)^{3-4\infty}$$
$$=\left(\frac{\infty(4\infty+5)}{\infty(4\infty+3)}\right)^{-4\infty}=\left(\frac{4\infty}{4\infty}\right)^{-4\infty} = 1^{-4\infty} = \boxed{1}$$
However, when I try to graph the function, I can't reliably get my answer due to precision limitations, and I feel that this method of calculating limits is less than ideal. How can I confirm that this is indeed the limit?
 [/INST] $$\lim_{n\to\infty}\left(\frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$$
$$=\lim_{n\to\infty}\left(\frac{4n^2+3n-10+(2n+4)}{4n^2+3n-10} \right)^{3-4n}$$
$$=\lim_{n\to\infty}\left(1+\frac{(2n+4)}{4n^2+3n-10} \right)^{3-4n}$$
From $\lim_{n\to\infty} \frac{2n+4}{4n^2+3n-10}=0$, 
$$\Rightarrow \lim_{n\to\infty}\left(1+\frac{(2n+4)}{4n^2+3n-10} \right)^{3-4n}$$
$$=\lim_{n\to\infty}\left(1+\frac{(2n+4)}{4n^2+3n-10} \right)^{\left(\frac{4n^2+3n-10}{2n+4}\right)\times \left(\frac{2n+4}{4n^2+3n-10}\right) \times 3-4n}$$
$$=\lim_{n\to\infty} e^{\frac{(2n+4)(3-4n)}{4n^2+3n-10}}$$
$$=e^{-2}$$
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