Calculate a lim $\lim_{x\to \infty } \left ( \frac{x^2+2}{x^2-4} \right)^{9x^2} $ $$
\lim_{x\to \infty } \left ( \frac{x^2+2}{x^2-4} \right)^{9x^2} 
$$
Can you help with it?
 A: Using the polynomial division algorithm, rewrite the base as
$$\frac {x^2 + 2}{x^2 - 4} = 1 + \frac 6{x^2 - 4} = 1 + \frac 1 {\frac {x^2 - 4}{6}}$$
We will make use of the standard limit:
$$\lim_{x \to +\infty} \left(1 + \frac 1x \right)^x = e$$
So:
$$\begin{align}
\lim_{x \to +\infty} \left(\frac {x^2 + 2}{x^2 - 4}\right)^{9x^2} &= \lim_{x \to +\infty} \left[\left(1 + \frac 1 {\frac {x^2 - 4}{6}}\right)^{\Large\frac {x^2 - 4}{6} \cdot \frac 6 {x^2 - 4}}\right]^{9x^2}=\\
&=\lim_{x \to +\infty} \left[\left(1 + \frac 1 {\frac {x^2 - 4}{6}}\right)^{\Large\frac {x^2 - 4}{6}}\right]^{\Large\frac {6\cdot9x^2}{x^2 - 4}}=\\
&=e^{54}
\end{align}$$
A: Hint: $\left(\dfrac{x^2+2}{x^2-4}\right)^{9x^2} = \left(\left(1+\dfrac{6}{x^2-4}\right)^{x^2-4}\right)^9\cdot \left(1+\dfrac{6}{x^2-4}\right)^{36}$
A: Hint
Let $$A=\left ( \frac{x^2+2}{x^2-4} \right)^{9x^2}$$ and take logarithms of both sides $$\log(A)=9x^2 \log\Big(\frac{x^2+2}{x^2-4}\Big)=9x^2\log\Big(\frac{1+\frac{2}{x^2}}{1-\frac{4}{x^2}}\Big)$$ $$\log(A)=9x^2 \Big[\log \left(1+\frac{2}{x^2}\right)-\log \left(1-\frac{4}{x^2}\right) \Big]$$ Now, use the fact that, for mall $y$, $$\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+O\left(y^4\right)$$ Apply that to each of the logarithms separately, replace $y$ by its expression as a function of $x$ (that is to say $\frac{2}{x^2}$ for the first and $\frac{-4}{x^2}$ for the second)  and simplify. 
I am sure that you can take from here.
