Let $X_1,X_2,\dots$ be iid random variables with mean zero and finite variance, and let $S_n = \sum_{k=1}^n X_k$. For $b>0$, find the limit $$\lim_{n\to\infty}\mathbb{P}\!\left(\left\lvert\frac{1}{\sqrt{n}}S_n\right\rvert<b\right)$$
I know to use central limit theorem to solve this, but I just don't know how to approach it.
The CLT indeed implies that $$\frac{S_n}{\sqrt{n}} \overset{d}\to N(0,σ^2)$$ where $σ^2$ denotes the finite variance of the $X_i$'s, i.e. $σ^2=Var(X_i)$, for $i\in \mathbb N$. So, \begin{align}P\left(\left|\frac{1}{\sqrt{n}}S_n\right|<b\right)&=P\left(-b< \frac{1}{\sqrt{n}}S_n<b\right)\\[0.3cm]&=P\left(\frac{1}{\sqrt{n}}S_n<b\right)-\left(\frac{1}{\sqrt{n}}S_n<-b\right)\\[0.3cm]&\to Φ(b/σ)-Φ(-b /σ)=2Φ(b/σ)-1\end{align} as $n\to +\infty$.