Help Finding the limit of the lognormal pdf about $0$

Question

I am trying to find the pdf of the $\operatorname{lognormal}(\mu, \sigma^2)$ distribution as the RV approaches $0$.

$f(x; \mu, \sigma) = \frac{1}{(2\pi)^{1/2} \sigma*x} \exp{(-\frac{1}{2}(\frac{\log(x)-\mu}{\sigma})^2)}$

I am absolutely lost as to how to actually do this, as several iterations of L'Hôpital's have done nothing to get rid of this pesky rogue x in the denominator.

Wolfram tells me that the limit converges to $0$, with no hints as to how this actually works.

Any insight would be really helpful here.

Answer 1

Let's consider the case where $x \to 0+$ since $x \in (0, +\infty)$

$$\lim_{x\to 0+} \frac{1}{x \sigma \sqrt{2 \pi}} e^{-\frac{(\ln(x)-\mu)^2}{2 \sigma^2}}$$

Also, since

$$\frac{1}{e^{\frac{(\ln(x)-\mu)^2}{2 \sigma^2}} \sigma \sqrt{2 \pi}} e^{-\frac{(\ln(x)-\mu)^2}{2 \sigma^2}} \leq \frac{1}{x \sigma \sqrt{2 \pi}} e^{-\frac{(\ln(x)-\mu)^2}{2 \sigma^2}} \leq \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(\ln(x)-\mu)^2}{2 \sigma^2}}$$

Simplifying...

$$\frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(\ln(x)-\mu)^2}{\sigma^2}} \leq \frac{1}{x \sigma \sqrt{2 \pi}} e^{-\frac{(\ln(x)-\mu)^2}{2 \sigma^2}} \leq \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(\ln(x)-\mu)^2}{2 \sigma^2}}$$

It is clear that

$$\lim_{x\to 0} \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(\ln(x)-\mu)^2}{\sigma^2}}=0$$

And

$$\lim_{x\to 0} \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(\ln(x)-\mu)^2}{2 \sigma^2}} = 0$$

Since the exponential functions decays to 0. Therefore, by the sandwich theorem, the middle term (which we want to solve) also decays to 0.

Answer 2

We seek an $x\to0^+$ one-sided limit. In terms of $y:=-\ln x$ it's$$\frac{1}{\sigma\sqrt{2\pi}}\exp\lim_{y\to\infty}\left(y-\frac{(y+\mu)^2}{2\sigma^2}\right).$$The exponent is a "sad" quadratic (i.e. its leading coefficient is negative), making the result $\frac{1}{\sigma\sqrt{2\pi}}\exp(-\infty)=0$.