Showing the limit of a function tending to zero I need to show that:
$$\lim_{x\to0}\frac {2\sigma^2(1-e^{-\theta x^2})} {x^2}=2\sigma^2\theta$$
By plugging in arbitrary values for the constants and trying different values for x I can see that this is true but I'm not sure how to show it formally. As x tends to 0, I know that the term in brackets will get closer to 0 too. I guess I need to somehow show that at values near zero, the term in brackets should be close to $\theta / x^2$?
 A: well this limit is indeterminant i.e. $\lim_{x\to 0}\frac{2\sigma^2(1-e^{-\theta x^2})}{x^2}=\frac{0}{0}$. use Lhopital's rule, and differentiate the top and bottom to find 
$$\lim_{x\to 0}\frac{2\sigma^2(1-e^{-\theta x^2})}{x^2}=\lim_{x\to 0}\frac{2\sigma^2(2\theta)xe^{-\theta x^2}}{2x}=\lim_{x\to 0}2\sigma^2\theta e^{-\theta x^2}=2\sigma^2\theta.$$
A: As an alternative approach, consider the following transformation:
$$\begin{align}\lim_{x\to 0}\frac {2\sigma^2(1-e^{-\theta x^2})}{x^2}&=2\sigma^2\lim_{x\to 0}\frac {1-\sum_{n=0}^\infty \frac {(-\theta x^2)^n}{n!}}{x^2}\\
&=2\sigma^2\lim_{x\to 0}\theta-\sum_{n=2}^\infty \frac {(-\theta)^n x^{2n-2}}{n!}\\
&=2\sigma^2\theta\end{align}$$
A: $$\lim_{x\to0}\frac {2\sigma^2(1-e^{-\theta x^2})} {x^2}=2\sigma^2\theta \frac{1-e^{-\theta x^2}}{\theta x^2}$$
After the substitution $t=-\theta x^2$ you get
$$2\sigma^2\theta \lim_{t\to0} \frac{1-e^{t}}{-t}
=2\sigma^2\theta \lim_{t\to0} \frac{e^{t}-1}{t} = 2\sigma^2\theta.$$
(In the last limit you can us L'Hospital, or just notice that this is precisely the definition of derivative of $e^t$ at 0, or use any other method to calculate this familiar limit. See Proving that $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$ and Show $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$.)
