Please prove that $\lim\limits_{n \to \infty} \int_0^3 \sqrt{\sin(n/x)+x+1}\,dx$ exists and evaluate it. Prove that $\displaystyle{\lim_{n \rightarrow \infty}} \int_0^3 \sqrt{\sin(n/x)+x+1}\,dx$ exists and evaluate it.
I'd love to take a calculus class in which such problems are assigned as homework. Not sure I would pass it, though.

First, let's consider the claim (made in another answer) that $\sin(n/x)\to 0$ as $n\to\infty$. Of course, this is false pointwise. But as a sequence in $L^\infty([0,3])$, $\sin(n/x)$ converges to $0$ in the weak* sense:  namely, for every $\varphi\in L^1([0,3])$ we have 
$$\lim_{n\to\infty} \int_0^3 \sin(n/x)\, \varphi(x)\,dx =0 \tag1$$
One way to prove (1) is to approximate $\varphi$ by a smooth function with compact support  in $(0,3)$, change the variable $y=1/x$, and integrate by parts:
$$
\int_{1/3}^\infty \sin(ny)\, y^{-2}\varphi(1/y)\,dy = \frac{1}{n} \int_{1/3}^\infty \cos (ny )\, (y^{-2}\varphi(1/y))'\,dy \to 0
\tag2$$ 
since the integrand is bounded in $L^1$ norm. 

If we did not have the square root, the story would end with
$$ \int_0^3 (\sin(n/x)+x+1)\,dx \to \int_0^3 (x+1)\,dx \tag3$$
by virtue of weak* convergence. But the square root changes everything: weak* limits do not commute with nonlinear operations such as squaring or taking root. We can scratch everything written so far.



Let's take a broader view: we are integrating the composition of continuous function $F(x,y)=\sqrt{y+x+1}$, defined on $[0,3] \times [-1,1]$, with the oscillatory function $y=\sin(n/x)$. Forgetting the particular form of $F$, what should the limit
$$\lim_{n\to\infty} \int_0^3 F(x,\sin(n/x))\,dx \tag4$$ 
be in terms of $F$? Being a positive linear functional of $F$, (4) must be of the form $\iint F(x,y)\,d\mu(x,y)$ for some positive measure $\mu$ on $[0,3] \times [-1,1]$. Let $\mu_n$ be the pushforward of the Lebesgue measure on $[0,3]$ under $x\mapsto (x,\sin(n/x))$. This is a measure of total mass $3$ supported on a wiggly curve. This measure $\mu$ will be the weak* limit (now in the sense of dual of $C(K)$) of $\mu_n$. A little computation involving $|\sin't|=\sqrt{1-\sin^2 t}$ reveals this limit: 
$$\mu(x,y)=\frac{1}{\pi\sqrt{1-y^2}}\,dx\,dy \tag5$$
Since this is calculus homework, I shouldn't have to justify (5) in detail. I'll only say that this is the measure you get by projecting arclength measure on a circle to its diameter, and normalizing. It's also the weight for which Chebyshev polynomials are orthogonal, for this very  reason.

Now we are almost done: the limit is equal to
$$\int_{-1}^1 \int_0^3 \frac{\sqrt{y+x+1}}{\pi \sqrt{1-y^2}} \,dx\,dy = \frac{2}{3\pi}\int_{-1}^1 \frac{(y+4)^{3/2}-(y+1)^{3/2}}{\sqrt{1-y^2}}\,dy \tag6$$
Because I think that C- will be enough for me, I'll just stick (6) into Maple: the answer is
$$\boxed{\dfrac{4}{9\pi} \left(16\sqrt{5}\,E\left(\sqrt{2/5}\right)-3\sqrt{5}\,K\left(\sqrt{2/5}\right)-4\sqrt{2}\right)} \tag7$$
where $K$ and $E$ are complete elliptic integrals of the first and second kind, respectively. Having cheated with Maple, I wouldn't want to lose even more points for not putting my answer in a box.

Numerically, (7) is about $4.5957$, which agrees with direct numeric computations of the given integral.



For comparison, $\int_0^3 \sqrt{x+1}\,dx=14/3\approx 4.6667$. The square root is concave, which means that averaging it by means of oscillatory function stuck inside makes the integral smaller. This is why the answer is strictly less than $14/3$.
As sine is continous everywhere, we do not need to worry about using any special rules. 

$\sin(n/x) \to 0$ as $n \to \infty$ and so

$\lim_{n \to \infty} \int^3_0 f(x) \mathrm{d} x \to \int^3_0 (x+1)^{1/2} \mathrm{d} x$ which just boils down to reversing the chain rule.