Prove that lim$_{n → \infty} \int_{0}^{3} \sqrt(sin \frac{x}{n} + x + 1) dx$ exists and evaluate it Prove that the following limits exist and evaluate them.
c) $\lim_{n \to \infty} \int_{0}^{3} \sqrt{\sin \frac{x}{n} + x + 1}\ \text dx$
I need to use the following theorem from analysis;
Suppose $f_n \to f$ uniformly on a closed interval [a,b]. If each $f_n$ is integrable on $[a,b]$, then so is $f$ and $\lim_{n \to \infty} \int_{a}^{b}f_n(x) \text dx = \int_{a}^{b}[\lim_{n \to \infty}f_n(x) ]\text dx $.
My attempt: Let $f_n = \sqrt{\sin \frac{x}{n} + x + 1}$. Recall |$\sin\frac{x}{n}| \leq 1$ for all x in [0,3]. 
First we try to find the $\lim_{n\to \infty} f_n = f(x)$ so it we can for every $\epsilon > 0$ there is an $N$ such that $n \geq N$ implies $|f_n(x) - f(x) | < \epsilon$
However, the limit does not exists when for  $f_n = \sqrt{\sin \frac{x}{n} + x + 1}$ because of sine. 
Can someone please help me? I would really appreciate it.
 A: Let $f_n(x)=\sqrt{\sin{\frac{x}{n}}+x+1}$ and $f(x)=\sqrt{x+1}$. Then $f_n\to f$ uniformly on $[0,3]$ (this requires proof but not too difficult) and then use your Uniform Convergence Theorem to get
$$
\lim_{n\to\infty}\int_0^3\sqrt{\sin{\frac{x}{n}}+x+1}\,dx=\int_0^3\lim_{n\to\infty}\sqrt{\sin{\frac{x}{n}}+x+1}\,dx=\int_0^3\sqrt{x+1}\,dx
$$
which you finish off with a quick $u-$ substitution.
UPDATE: I just reread your post and I see the problem is the uniform convergence part of the proof. Recall that for sufficiently large $n$ we have $\sin\frac{x}{n}\approx\frac{x}{n}$.
UPDATE 2: Rather than an $\varepsilon-\delta$ proof, another way we can demonstrate uniform convergence is if
$$
\sup_{x\in X}|f_n(x)-f(x)|\to 0\,\,\text{ as }\,\,n\to\infty.
$$
So
$$
\begin{align*}
\sup_{x\in X}|f_n(x)-f(x)|&=\sup_{x\in [0,3]}\left\lvert\sqrt{\sin{\frac{x}{n}}+x+1}-\sqrt{x+1}\,\right\rvert\\
&=\sqrt{\sin{\frac{3}{n}}+3+1}-\sqrt{3+1}\to 0\,\,\text{ as }\,\,n\to\infty.
\end{align*}
$$
A: I am not sure that this could be the answer you expect; so, please forgive me if I am off-topic.
When $n$ is large, we can approximate $\sin(x)$ by its Taylor expansion built at $x=0$. The problem is that only the first term of the expansion can be used (otherwise we should bump on nasty elliptic integrals).
$$I_n=\int_{0}^{3} \sqrt{\sin (\frac{x}{n}) + x + 1}\, dx\approx \int_{0}^{3} \sqrt{\frac{x}{n} + x + 1}\, dx=\frac{2 \left(\sqrt{\frac{3}{n}+4}\, (4 n+3)-n\right)}{3 (n+1)}$$ the limit of which being $\frac{14}3$ which is, as expected, the value of $\int_{0}^{3} \sqrt{ x + 1}\, dx$.
Comparing with the results of numerical integration (given in parentheses), the match is quite correct as shown below for a few values on $n$.
$$I_{5}=4.92550\cdots (4.91865)$$
$$I_{10}=4.79798\cdots (4.79709)$$
$$I_{15}=4.75465\cdots (4.75438)$$
$$I_{20}=4.73282\cdots (4.73271)$$
A Taylor expansion of the approximation gives $$I_n \approx \frac{14}{3}+\frac{4}{3 n}-\frac{5}{24 n^2}+O\left(\left(\frac{1}{n}\right)^3\right)$$ showing how the limit is reached.
Curve fitting the results of the numerical integration for the range $5\leq n\leq 100$ leads to $$I_n=\frac{14}{3}+\frac{1.34057}{n}-\frac{0.389332}{n^2}$$ which is extremely close.
