Convolution and CLT

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I was revising for probability exam and this question caught my eye.

Let $f:R\to R$ be a non-negative function s.t $\int_R f(x)dx=1$ , $\int_Rxf(x)dx=0$ and $\int_Rx^2f(x)dx=25$

Find $$\lim_{n\to\infty}\int_1^\sqrt n \underbrace{f*f*\dots*f(x)}_{n-1\text{ times}}dx$$

Attempted solution:

Let $\{X_i\}_{i\in N}$ be a sequence of $i.i.d$ such that $X_i$ density function is $f(x)$. it follows that $EX_1 = 0, varX_1 =25$

$$\lim_{n\to\infty}\int_1^\sqrt n \underbrace{f*f*\dots*f(x)}_{n\text{ times}}dx=\lim_{n\to\infty}P\left(1\le \sum_{k=1}^{n}X_k\le\frac{1}{\sqrt{n}}\right)=\lim_{n\to\infty}P\left(\frac{1}{\sqrt{n}5}\le \frac{\sum_{k=1}^{n}X_k-0n}{5\sqrt{n}}\le\frac{1}{5}\right)$$

By the CLT, that probability is equal to $P(0\le Z\le\frac{1}{5})$ where $Z\sim N(0,1)$ and therefore, the answer is $\phi(1/5)-(1-\phi(0))$

The question is, am I allowed to assume the left hand of the probability tends to zero? The version of the CTL that I learnt had $P(Z<k)$ where $k$ is a scalar, the left hand side is a function of $n$, does it matter or is it correct regardless?

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Yes you can. Let assume we have a sequence of random real variables $(X_i)_{i \in \mathbb{N}}$ which converge on low to a random variable $Z$ and let assume we have a sequence $(a_i)_{i \in \mathbb{N}}$ which converge to a real value $a$. By Slutsky Lemma the sequence $(X_i-a_i)$ converge on low to $Z - a$ and so $P(0\le X_i - a_i)$ converge to $P(0 \le Z - a)$.