I need see another way to evaluate this limits without using L'Hôpital's rule (H-L). Problem:
Determine the folowing limit:
$$\displaystyle\lim\limits_{a\to 0}\frac{\delta_{a}+\delta_{-a}-2\delta}{a^{2}}$$
where $\delta_{a} $ is the Dirac distribution at the point $a$.
Answer is $\delta''$. My try using definition:
Let
$$I_{a}=\frac{\delta_{a}+\delta_{-a}-2\delta}{a^{2}}$$
Then:
$$\langle I_{a},\varphi \rangle =\langle \delta , \frac{\varphi (a)+\varphi (-a)-2\varphi (0)}{a^{2}} \rangle$$ Now lim of $I_{a}$
$$\lim\limits_{a\to 0} \langle I_{a},\varphi \rangle = \langle \delta ,\lim\limits_{a\to 0} \frac{\varphi (a)+\varphi (-a)-2\varphi (0)}{a^{2}} \rangle$$ because distribution continuous; now we find:
$$\lim\limits_{a\to 0} \frac{\varphi (a)+\varphi (-a)-2\varphi (0)}{a^{2}} =\lim\limits_{a\to 0}\frac{\varphi '(a)+\varphi ' (-a)}{2a}= \lim\limits_{a\to 0} \frac{\varphi ''(a)+\varphi '' (-a)}{2} =\varphi ''(0)$$
then the result.
If you know a simple method, share in this company. Thanks!
So this seems mostly fine except I think your expression for the very last part of your equality and $\langle I_a, \varphi\rangle$ should really just be
$$ \frac{\varphi(a) + \varphi(-a) - 2\varphi(0)}{a^2}. $$
You've already taken care of the $\delta$ by definition, no need to keep it around. Then
$$\lim_{a\to 0} \langle I_a, \varphi\rangle = \varphi''(0)$$
and so you can associate $\lim_{a\to 0} \frac{\delta_a + \delta_{-a} - 2\delta_0}{a^2}$ with $\delta''$ with usual distribution theory duality arguments. It is mistaken to write
$$\lim_{a\to 0}\frac{\varphi'(a) - \varphi'(-a)}{2a} = \delta''.$$
(I think you have a sign error.) This is actually equal to $\varphi''(0)$ (per the above) which then allows you to connect back to $\delta''$—but these are not the same thing. One is a compactly supported smooth function; the other is a tempered distribution.