Finding the limit of function - exponential one Find the value of $\displaystyle \lim_{x \rightarrow 0}\left(\frac{1+5x^2}{1+3x^2}\right)^{\frac{1}{\large {x^2}}}$
We can write this limit function as : 
$$\lim_{x \rightarrow 0}\left(1+ \frac{2x^2}{1+3x^2}\right)^{\frac{1}{\large{x^2}}}$$
Please guide further how to proceed in such limit..
 A: Write it as
$$\dfrac{(1+5x^2)^{1/x^2}}{(1+3x^2)^{1/x^2}}$$ and recall that $$\lim_{y \to 0} (1+ay)^{1/y} = e^a$$ to conclude what you want.
A: $$\lim_{x \rightarrow 0}\left(1+ \frac{2x^2}{1+3x^2}\right)^{\frac{1}{x^2}}$$
$$=\lim_{x \rightarrow 0}\left(\left(1+ \frac{2x^2}{1+3x^2}\right)^{\huge{\frac{1+3x^2}{2x^2}}}\right)^{\huge{\frac2{1+3‌​x^2}}}$$
$$=\left(\lim_{\frac{2x^2}{1+3x^2} \rightarrow 0}\left(1+ \frac{2x^2}{1+3x^2}\right)^{\huge{\frac{1+3x^2}{2x^2}}}\right)^{\huge\lim_{x \rightarrow 0}{\frac2{1+3‌​x^2}}}=e^2$$
as $x\to0\implies \frac{2x^2}{1+3x^2}\to0$
A: Let $n = \displaystyle\frac{1}{x^{2}}$:
$\displaystyle \lim_{x \rightarrow 0}\left(\frac{1+5x^2}{1+3x^2}\right)^{\frac{1}{\large {x^2}}}$ = $\displaystyle\lim_{n \rightarrow \infty}\left(\frac{n + 5}{n + 3}\right)^{n}$ = $\displaystyle\lim_{n \rightarrow \infty}\left(1 + \frac{2}{n + 3}\right)^{n}$ = $\displaystyle\lim_{n \rightarrow \infty}\left(1 + \frac{2}{n}\right)^{n}$ = $e^{2}$ 
by definition of $e$.
A: Based on the fact that
$$\forall\,a\in\Bbb R\;,\;\;\lim_{x\to x_0}\left(1+\frac{a}{f(x)}\right)^{f(x)}=e^a\;,\;\;\text{whenever $\,f\,$ is a function s.t.}\;\;f(x)\xrightarrow[x\to x_0]{}\infty$$
we get:
$$\left(1+ \frac{2x^2}{1+3x^2}\right)^{\frac{1}{x^2}}=\left(1+ \frac{2}{3+\frac{1}{x^2}}\right)^{3+\frac{1}{x^2}}\left(1+ \frac{2}{3+\frac{1}{x^2}}\right)^{-3}\xrightarrow[x\to0]{}e^2\cdot  1=e^2$$
