<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] Let us suppose that you want to compute accurately the value of the expression for large values of $n$ and not only the limit.
$$a_n=\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}=\left(1+\frac{2}{4 n-5}\right)^{3-4 n}$$ Take logarithms
$$\log(a_n)=(3-4n) \log\left(1+\frac{2}{4 n-5}\right)$$ and use the Taylor expansion
$$\log(1+\epsilon)=\epsilon -\frac{\epsilon ^2}{2}+\frac{\epsilon ^3}{3}-\frac{\epsilon
   ^4}{4}+O\left(\epsilon ^5\right)$$
Make $\epsilon=\frac{2}{4 n-5}$ and continue with Taylor expansion (or long division) to get
$$\log(a_n)=(3-4n)\left(\frac{1}{2 n}+\frac{1}{2 n^2}+\frac{49}{96 n^3}+\frac{17}{32
   n^4}+O\left(\frac{1}{n^5}\right)\right)$$ that is to say
$$\log(a_n)=-2-\frac{1}{2 n}-\frac{13}{24 n^2}-\frac{19}{32
   n^3}+O\left(\frac{1}{n^4}\right)$$ COntinue with Taylor using
$$a_n=e^{\log(a_n)}=\frac 1{e^2}\left(1-\frac{1}{2 n}-\frac{5}{12 n^2}-\frac{11}{32
   n^3} \right)+O\left(\frac{1}{n^4}\right)$$
Try with $n=5$ (which is quite far away from $\infty$. The exact value is
$$a_5=\frac{98526125335693359375}{827240261886336764177}\approx 0.119102$$ while the above truncated expression gives
$$\frac{10567}{12000 e^2}\approx 0.119174$$
</s>