Derivation of Stirling approximation from CLT In Casella Berger exercise 5.35, we have a sequence of sample means from the exponential $exp(1)$. Then, we can derive:
$$P\Bigg(\frac{\overline{X}_n - 1}{1/\sqrt{n}} \leq x\Bigg) \rightarrow P(Z \leq x)$$
where $Z \sim n(0, 1)$.
Now, the author suggests to show the Stirling approximation by taking the derivative on both sides. In short, he suggests:
$$\frac{d}{dx}P\Bigg(\frac{\overline{X}_n - 1}{1/\sqrt{n}} \leq x\Bigg) \approx \frac{d}{dx}P(Z \leq x)$$
from where the Stirling approximation follows.
Why are we justified doing this? Wiki article on weak convergence says:
"In general, convergence in distribution does not imply that the sequence of corresponding probability density functions will also converge."
Why is taking the derivative justified according to the author?
If it is allowed, can be it shown in some simple way without measure theory? Or link to the relevant theorems.
 A: Let $f_n$ be the pdf of $Z_n=\sqrt{n}(\sum_i X_i/n-1)$ and $f$ be the pdf of a standard Gaussian. Then the exercise leads you to
$$f_n(0)=\frac{1}{n!}\left(\frac{n}{e}\right)^n \sqrt{n}$$
As you mention, from there we cannot simply use convergence in distribution to conclude that $f_n(0)\to f(0)=1/\sqrt{2\pi}$. Here is an argument based on characteristic functions to bypass this issue.

Let $\Phi_n(t)=e^{-it\sqrt{n}}(1-it/\sqrt{n})^{-n}$ be the characteristic function of $Z_n$ (note that it converges pointwise to $e^{-t^2/2}$, either by the CLT or by direct computation). Then,
$$|\Phi_n(t)|= \left| 1-\frac{it}{\sqrt{n}}\right|^{-n}=\left(1+\frac{t^2}{n}\right)^{-n/2}=\exp\left(-\frac{n}{2}\ln\left(1+\frac{t^2}{n}\right)\right)$$
Now it is just a simple exercise in real analysis to show that $|\Phi_n(t)|$ is uniformly dominated by an integrable function (treat the cases $t^2\le n$ and $t^2>n$ separately). Thus, from an inversion theorem and dominated convergence,
$$f_n(0)=\frac{1}{2\pi}\int_{-\infty}^{\infty} \Phi_n(t)\,dt\to 1/\sqrt{2\pi}.$$
