Showing the limit of a function tending to zero

Question

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 $\theta / x^2$?

Consider writing $e^{-\theta x^2}$ in summation form to see what happens... — abiessu, Sep 25 '14 at 18:53

Ellya · Accepted Answer · 2014-09-25T19:00:31.397

3

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 $$\lim_{x\to 0}\frac{2\sigma^2(1-e^{-\theta x^2})}{x^2}=\lim_{x\to 0}\frac{2\sigma^2(2\theta)xe^{-\theta x^2}}{2x}=\lim_{x\to 0}2\sigma^2\theta e^{-\theta x^2}=2\sigma^2\theta.$$

edited Sep 25 '14 at 19:00

answered Sep 25 '14 at 18:53

Ellya

11,783

Ah, I've not heard of this rule. Thanks very much! – user123965 Sep 25 '14 at 18:56
I think the derivative of the expression's numerator is $$;2\sigma^2\cdot 2\theta x e^{-\theta x^2}=4x\theta e^{-\theta x^2};$$ – Timbuc Sep 25 '14 at 18:59
@Timbuc oh yeah sorry, still works out the same I will edit :) – Ellya Sep 25 '14 at 18:59
Indeed it works just the same because $;e^{-\theta x^2}\xrightarrow[x\to 0]{}1;$ ,@ellya . +1 – Timbuc Sep 25 '14 at 19:00

score 1 · Answer 2 · answered Sep 25 '14 at 19:06

As an alternative approach, consider the following transformation:

$$\begin{align}\lim_{x\to 0}\frac {2\sigma^2(1-e^{-\theta x^2})}{x^2}&=2\sigma^2\lim_{x\to 0}\frac {1-\sum_{n=0}^\infty \frac {(-\theta x^2)^n}{n!}}{x^2}\\ &=2\sigma^2\lim_{x\to 0}\theta-\sum_{n=2}^\infty \frac {(-\theta)^n x^{2n-2}}{n!}\\ &=2\sigma^2\theta\end{align}$$

score 0 · Answer 3 · edited Apr 13 '17 at 12:20

$$\lim_{x\to0}\frac {2\sigma^2(1-e^{-\theta x^2})} {x^2}=2\sigma^2\theta \frac{1-e^{-\theta x^2}}{\theta x^2}$$

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 $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$.)

Showing the limit of a function tending to zero

3 Answers3