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?

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:
$$\begin{align}
\lim_{x \to +\infty} \left(\frac {x^2 + 2}{x^2 - 4}\right)^{9x^2} &= \lim_{x \to +\infty} \left[\left(1 + \frac 1 {\frac {x^2 - 4}{6}}\right)^{\Large\frac {x^2 - 4}{6} \cdot \frac 6 {x^2 - 4}}\right]^{9x^2}=\\
&=\lim_{x \to +\infty} \left[\left(1 + \frac 1 {\frac {x^2 - 4}{6}}\right)^{\Large\frac {x^2 - 4}{6}}\right]^{\Large\frac {6\cdot9x^2}{x^2 - 4}}=\\
&=e^{54}
\end{align}$$
