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?
 A: 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
A: $$\lim_{n\to\infty}\left(\frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$$
$$=\lim_{n\to\infty}\left(\frac{4n^2+3n-10+(2n+4)}{4n^2+3n-10} \right)^{3-4n}$$
$$=\lim_{n\to\infty}\left(1+\frac{(2n+4)}{4n^2+3n-10} \right)^{3-4n}$$
From $\lim_{n\to\infty} \frac{2n+4}{4n^2+3n-10}=0$, 
$$\Rightarrow \lim_{n\to\infty}\left(1+\frac{(2n+4)}{4n^2+3n-10} \right)^{3-4n}$$
$$=\lim_{n\to\infty}\left(1+\frac{(2n+4)}{4n^2+3n-10} \right)^{\left(\frac{4n^2+3n-10}{2n+4}\right)\times \left(\frac{2n+4}{4n^2+3n-10}\right) \times 3-4n}$$
$$=\lim_{n\to\infty} e^{\frac{(2n+4)(3-4n)}{4n^2+3n-10}}$$
$$=e^{-2}$$
A: In addition to all the other answers, that give you the correct solution,
let me phrase in no uncertain terms that your solution attempt is very misguided and will lead you to wrong results in many cases.
There are reasons why you almost never see a matematician use thy $\infty$-symbol in a formula like $4\infty+5$. The main reason is that many properties about numbers that we are tought and use intuitively are not true if you think about limits reaching an actual 'number' $\infty$. 
If you have two sequences $a_n$ and $b_n$ with the limits $$\lim_{n\to \infty}a_n=\infty,\quad \lim_{n\to \infty}b_n=\infty,$$
then 
$$\lim_{n\to \infty}(a_n-b_n) \text{ could be anything from }+\infty, 2019, 0, -\infty, \text {any other real number or not existing at all.}$$
 You cannot model that with the formula $\lim_{n\to \infty}(a_n-b_n)=\infty - \infty$. Once you use $\infty$ as an actual number, you have lost all the information how the series converges to $\infty$, which means you cannot tell which of the cases I described is actually correct for two given series $a_n$ and $b_n$.
The same is true for $\lim_{n\to \infty}\frac{a_n}{b_n}$, which you used in your formulas. 
The way you handled this shows that you have some correct intuition about limits, but unfortunately at the crucial moment, when you are dealing with
$$\left(\frac{4\infty}{4\infty}\right)^{-4\infty}$$
you are now a victim of what you did before, because you have now no idea how either of the terms
$$ \frac{4n^2+5n-6}{4n^2+3n-10} \text{ and } 3-4n$$
converges to $1$ or $-\infty$, resp., which is important to know (if you look at the real solutions).
A: 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
$$\frac{10567}{12000 e^2}\approx 0.119174$$
