Prove that lim$_{n → \infty} \int_{0}^{3} \sqrt(sin \frac{x}{n} + x + 1) dx$ exists and evaluate it

Question

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.

Answer 1

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 $$ \begin{align*} \sup_{x\in X}|f_n(x)-f(x)|&=\sup_{x\in [0,3]}\left\lvert\sqrt{\sin{\frac{x}{n}}+x+1}-\sqrt{x+1}\,\right\rvert\\ &=\sqrt{\sin{\frac{3}{n}}+3+1}-\sqrt{3+1}\to 0\,\,\text{ as }\,\,n\to\infty. \end{align*} $$

Answer 2

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.