How to show that a lognormal distribution approaches normality for a small coefficient of variation? I note that the lognormal distribution approaches a normal for a small coefficient of variation, but I'm having a hard time showing this mathematically. Can someone help?
 A: This is a special case of a general phenomenon; and analyzing the generalization enables us avoid almost all details of Lognormal distributions.
Here's the sketch of the general idea.  Suppose $Z$ is any random variable (in the application it will have a Normal distribution) and $g$ is a real-valued function differentiable in a neighborhood $\mathscr U$ of $0$ with nonzero derivative there, $g^\prime(0)\ne 0.$ (In the application $g$ will be an exponential function).  By definition, this means
$$g(z) = g(0) + g^\prime(0)z + zh(z)$$
where $\lim_{z\to 0} h(z) = 0.$
The general claim is that

As $\sigma\to 0,$ the distribution of a suitably normalized version of $Y = g(\sigma Z)$ approaches the distribution of $Z.$

In the application, $\sigma$ is directly related to the coefficient of variation (CV) and $Y$ has a Lognormal distribution.  All Lognormal distributions arise in this way.
The "suitable normalization" must be
$$\frac{g(\sigma Z) - g(0)}{\sigma g^\prime(0)} = Z + \sigma\, (Z h(\sigma Z)).$$
The remainder term on the right cannot wholly be controlled by shrinking $\sigma,$ because $Z$ can be arbitrarily large, leaving us unable to say anything about the magnitude of $h(\sigma Z),$ no matter what (nonzero) value we use for $\sigma.$  However, to high probability we can approximate $Z$ by a bounded variable $Z_K$ in which $Z$ is replaced by some bounded quantity whenever $|Z|\gt K.$ When $K$ is large enough, the distribution of $Z_K$ is essentially that of $Z.$  By then making $|\sigma|$ suitably small, $\sigma Z_K$ lies in the arbitrarily small interval $[-|\sigma|K, |\sigma|K] \subset \mathscr U.$  The differentiability of $g$ at $0$ implies that within this small interval, $g$ is well approximated by an invertible linear function: that is, it is just shifting $Z_K$ (by $g(0)$) and rescaling it (by $\sigma g^\prime(0)$).  Consequently, undoing this shifting and rescaling recovers $Z$ to an arbitrarily good approximation with arbitrarily high probability, QED.

All lognormal distributions arise as the distributions of variables $Y_{\mu,\sigma} = \exp(\sigma Z + \mu)$ when $Z$ has a standard Normal distribution with distribution function $\Phi.$  The value of $|\sigma|$ determines the coefficient of variation and grows small as the CV grows small.  What we want to find, then, is a limiting distribution for $Y_{\mu,\sigma}$ as $|\sigma| \to 0.$
Because the CV reflects the spread of $Y$ relative to its mean, $Y_{\mu,\sigma}$ approaches a constant.  At left, to illustrate, is a histogram of 100,000 realizations of $Y_{-1/2, 1/100}.$

More insight can be had by suitably Normalizing $Y.$  Let us consider what happens to $$\left(e^{-\mu}\,Y-1\right)/\sigma = \left(e^{\sigma Z} - 1\right)/\sigma$$
as $|\sigma|\to 0.$  (Its histogram is shown in the right panel of the illustration.)  To do that, let's estimate its distribution function,
$$\Pr\left(\left(e^{\sigma Z}-1\right)/\sigma \le t\right).$$
Using the Maclaurin series for the exponential, and taking $|\sigma|\le 1,$ it is immediate that
$$\bigg| \frac{e^{\sigma Z}-1}{\sigma} - Z\bigg| = |\sigma|\,\bigg| \frac{Z^2}{2!} + \frac{\sigma Z^3}{3!} + \cdots\bigg| \le |\sigma|\,e^{|Z|}$$
for all $Z.$  Let $\epsilon \gt 0$ be very small, ensuring there is $z_\epsilon \gt 0$ for which $\Phi(-z_\epsilon) \le \epsilon/4,$ meaning $$\Pr(|Z| \ge z_\epsilon) \le \epsilon/2.$$
Let
$$0 \lt \sigma \le \frac{\epsilon}{2}\,\exp(z_\epsilon),$$
thereby guaranteeing
$$Z - \frac{\epsilon}{2} \le \frac{e^{\sigma Z}-1}{\sigma} \le Z + \frac{\epsilon}{2}\tag{*}$$
provided $-z_\epsilon \le Z \le z_\epsilon.$
Since $z_\epsilon\to\infty$ as $\epsilon\to 0,$ this shows the distribution functions of $Z$ and $(e^{\sigma Z}-1)/\sigma$ are arbitrarily close provided $|\sigma|$ is sufficiently small. Consequently, the limiting distribution of the latter is the distribution function of $Z.$  That's your result:

As $\sigma$ approaches $0,$ the distribution of the standardized variable $(e^{-\mu}Y_{\mu,\sigma} - 1)/\sigma)$ approaches the distribution of $Z.$

As a further illustration, here is a simulation of 100,000 Poisson$(3)$ values for $Z:$

A close look at the right panel shows the histogram begins to deviate from a Poisson probability distribution for values exceeding $6$ or so: the bars, which in the limit will be located only at integral values, appear to be sliding a bit to the left of those integers.  This is what $(*)$ means: for small finite values of $\sigma,$ both the heights and the positions of values on the graph of the distribution function might be displaced a little, as limited by $\epsilon/2.$  But as $|\sigma|$ decreases, the graph eventually settles down to the distribution function of $Z$ itself.
