Why is this definition of the Central Limit Theorem not incorrect? I found the following definition of the Central Limit Theorem from book Probability and Statistics by Degroot (also from Wikipedia). It simply states the CLT as
$$ \lim_{n\to\infty} \Pr\bigg[\frac{\bar{X}_n-\mu}{\sigma/\sqrt n} \le z\bigg] = \Phi(z), \quad \cdots (1)$$
where $\mu$ is population mean, $\sigma$ is population standard deviation and 
$\bar{X}_n= \frac{1}{n}\sum_{i=i}^{n}X_i = \frac{1}{n} \left( X_1+X_2+ \cdots+X_n \right)$.
My concern is if $n \to \infty$ then why $\bar{X}$ would follow a nondegenerate distribution ? The law of large number states that
$$
    \Pr\!\left( \lim_{n\to\infty}\bar{X}_n = \mu \right) = 1$$
 If this is true then both numerator and denominator in equation 1 will converge to zero as $n \to \infty$ and $\bar{X}$ would become a constant instead of following a specific distribution. 
 A: *

*$\bar{X}$ is defined as $\frac{1}{n}\sum_{i=1}^nX_i$, your definition misses an average, and summation should start at $i=1$.

*$\mu$ is the population mean, not the sample mean ($\bar{X}$ is). Likewise, $\sigma$ is the population standard deviation.

*The crucial difference to the LLN is that the difference between $\bar{X}$ and $\mu$ (which indeed vanishes by the LLN) is scaled by $\sqrt{n}$, which diverges. So rewrite (1) as $\sqrt{n}(\bar{X}-\mu)/\sigma$, and it turns out (a proper proof would be too long here) that this product of two things, one of which tends to 0 and the other to infinity indeed (under suitable assumptions) still has a distribution asymptotically.
A: Note that in your expression
$$ \lim_{n\to\infty} \Pr\bigg[\frac{\bar{X}_n-\mu}{\sigma/\sqrt n} \le z\bigg] $$
There is nowhere a reference to $\lim_{n\to\infty}\bar{X}_n$. It doesn't matter what this last part converges to - you're working with a different expression. It seems you were trying to do something like
$$ \lim_{n\to\infty} \Pr\bigg[\frac{\bar{X}_n-\mu}{\sigma/\sqrt n} \le z\bigg] = \Pr\bigg[\frac{\lim_{n\to\infty}\bar{X}_n-\mu}{\sigma/\sqrt n} \le z\bigg] = \Pr\bigg[\frac{\mu-\mu}{\sigma/\sqrt n} \le z\bigg] = \mathrm{Pr}[0\le z]$$
But you can't do that, just like in normal calculus you can't do something like
$$\lim_{t\to\infty}\left(\frac1t\cdot t\right)=\left(\lim_{t\to\infty}\frac1t\right)t=0.$$
In reality, it's true that as $n$ increases, the difference $\bar{X}_n-\mu$ becomes smaller, but you also multiply by $\sqrt{n}$ which gets larger and offsets this. The combined effect is that as $n$ increases, $\frac{\bar{X}_n-\mu}{\sigma/\sqrt n}$ becomes closer to a standard normal distribution.
