<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] Hint:
$$\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)=\frac{(n+2)(4n-3)}{(n+2)(4n-5)}=\frac{4n-3}{4n-5}=\frac{1}{(1-\frac{2}{4n-3})}=(1-\frac{2}{4n-3})^{-1}$$
so
$$\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}=(1-\frac{2}{4n-3})^{4n-3}$$
then use
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