Evaluate $\lim_{x\to \infty} \left( \cfrac{2x^2-1}{2x^2+3} \right)^{4x^2+2}$ 
I'm stuck in evaluating the following limit. $$\lim_{x\to \infty} \left(\cfrac{2x^2-1}{2x^2+3}\right)^{4x^2+2} $$ 

This is what I've tried yet.
Let : $2x^2 + 1 = g(x)$ ; Then we've : $2x^2 - 1 = g(x) - 2 $ and $2x^2 + 3 = g(x) + 2 $ 
Therefore, the limit given can be written as: $$\lim_{x\to \infty} \left( \cfrac{g(x)-2}{g(x) + 2} \right)^{2g(x)}$$
Now, as $x \to \infty$ then $g(x) \to \infty$ .
Therefore, we've: $$\lim_{g(x) \to \infty} \left(\cfrac{g(x) - 2}{g(x)+2} \right)^{2g(x)} $$
Although, I know that this idea seems to be really dumb as far as proper technique for this individual question is concerned. But, I can't think of any other technique right now. 
Any help will be greatly appreciated. 
 A: Rewrite it as
$$
\left(1-\frac{4}{2x^2+3}\right)^{2(2x^2+3)-4}
$$
Call $n=2x^2+3$, then
$$
\lim_{n\to \infty}\left(1-\frac{4}{n}\right)^{2n-4}=e^{-8}.
$$
A: Taking log $$\lim_{x\rightarrow\infty}\frac{\log\left(\frac{2x^{2}-1}{2x^{2}+3}\right)}{1/\left(4x^{2}+2\right)}
 $$ and using L'Hopital's rule, we get $$\lim_{x\rightarrow\infty}-\frac{8\left(2x^{2}+1\right)^{2}}{\left(2x^{2}-1\right)\left(2x^{2}+3\right)}=-8
 $$ hence $$\lim_{x\rightarrow\infty}\left(\frac{2x^{2}-1}{2x^{2}+3}\right)^{4x^{2}+2}=e^{-8}.
 $$
A: Hint
Let $$A= \left(\cfrac{g - 2}{g+2} \right)^{2g}$$ and take logarithms $$\log(A)=2g\log\left(\cfrac{g - 2}{g+2} \right)=2g\log\left(\cfrac{1 -\frac 2 g}{1 +\frac 2 g} \right)=2g\Big( \log(1 -\frac 2 g)-\log(1 +\frac 2 g)\Big)$$ Now, use the development of $\log(1+x)$ when $x$ is small and replace successively $x$ by $ -\frac 2 g$ and $-\frac 2 g$.
After simplification, you should get $$\log(A)=-8-\frac{32}{3 g^2}+O\left(\left(\frac{1}{g}\right)^4\right)$$ and then $$A=\frac{1}{e^8}-\frac{32}{3 e^8 g^2}+O\left(\left(\frac{1}{g}\right)^4\right)$$ which reveals not only the limit but also how it is approached.
