Evaluate $\lim_{x\rightarrow 0}\frac{1}{\sqrt{x}}\exp\left[-\frac{a^2}{x}\right]$ I am interested in the following limit:
$$\lim_{x\rightarrow 0}\frac{1}{\sqrt{x}}\exp\left[-\frac{a^2}{x}\right]$$
Does this limit exist for real $a$?
Edit: I am only interested in the case when $x$ is non-negative. Thanks for reminding.
 A: Assuming you mean $\lim_{x \to 0^+}$, sure: introduce $u=1/x$, then you have
$$\lim_{u \to \infty} u^{1/2} e^{-a^2 u}$$
which is $0$ (for instance use L'Hopital's rule to prove it).
If you mean what you wrote, think about $x<0$.
A: Let's make some use of the theory of distributions.
First of all, let's substitute $x \to y^2$ so you have
$$\lim_{y\to 0}\ \frac{1}{y}e^{-a^2/y^2}$$
Considering that $y$ goes to zero, I avoided to write $|y|$.
Now, there is a Theorem which states that if a function $u(x) \in L^1$ and if $u_{\lambda}(x) = \lambda u(\lambda x)$ then
$$\lim_{\lambda\to +\infty} u_{\lambda}(x) = C\delta(x)$$
where $\delta(x)$ is Dirac Delta Function and $C = \int_{-\infty}^{+\infty}\ u(x)\ dx$
This works in our case: $$\frac{1}{y}e^{-a^2/y^2} = u_{\lambda}(a) = \lambda u(\lambda a)$$ where $$\lambda = \frac{1}{y}$$ and $$u(a) = e^{-a^2}$$
Thus, having the well known integral
$$\int_{-\infty}^{+\infty}\ e^{-a^2}\ da = \sqrt{\pi} = C$$
we get
$$\lim_{x\to 0} \frac{1}{\sqrt{x}}e^{-a^2/x} = \lim_{y\to 0} \frac{1}{y}e^{-a^2/y^2} = \lim_{\lambda\to +\infty} \lambda u(\lambda a) = C\delta(a) = \sqrt{\pi}\delta(a)$$
A: Hint: what happens when $x$ is negative?
A: The left hand limit is not defined because of the square root in the denominator. So, you just need to check the value of the function at $f(0^+)$ and $f(0)$. If these two are equal, then limit is defined at $x=0$.
