Limit of 0/0 type

Question

I would like to calculate the limit \begin{equation} \lim_{h\rightarrow0}\frac{h}{1-\frac{1}{\sqrt{2\pi h}} \int_{-d}^{d}\exp(-\frac{x^2}{2h})dx} \end{equation} It is of the 0/0 type. First, I change the limit into the following form \begin{equation} \lim_{h\rightarrow0}\frac{h}{1-\frac{1}{\sqrt{2\pi}} \int_{-d/\sqrt{h}}^{d/\sqrt{h}}\exp(-\frac{x^2}{2})dx} \end{equation}

Then, I use the L'Hospital's Rule once, it becomes
\begin{equation} \lim_{h\rightarrow0}\frac{h^{\frac{3}{2}}}{\frac{1}{\sqrt{2\pi}} \exp(-\frac{d^2}{2h})} \end{equation}

I can keep using the L'Hospital's Rule. But it is always the 0/0 type without end.

Answer 1

Your final expression is $$\sqrt{2\pi} \lim_{h \to 0} \exp\left(\frac{d^2}{2h}\right) h^{3/2}$$

Take the logarithm of the thing in the limit: it's $\dfrac{d^2}{2h} + \frac{3}{2} \log(h) = \frac{1}{h} \left(\dfrac{d^2}{2} + \frac{3}{2} h \log(h)\right)$.

Since $x \log(x) \to 0$ as $x \to 0$, the bracket clearly tends to $\dfrac{d^2}{2}$, so the logarithm of your original limit is $\lim_{h \to 0} \frac{d^2}{2h}$, which is $\infty$.

Answer 2

Alternative approach: using the well-known asymptotic approximation $\text{Erfc}(z)\sim \frac{e^{-z^2}}{z\sqrt{\pi}}$ we get

$$ \lim_{h\to 0^+}\frac{h}{\text{Erfc}\left(\frac{d}{\sqrt{2h}}\right)}=\lim_{h\to 0^+}hd\sqrt{2\pi h} e^{\frac{d^2}{2h}} = +\infty.$$