Help evaluating $\lim\limits_{h\to0}\int_{\frac{\pi}{4}}^{\frac{\pi}{4}+h}\frac{\sin(x)}{x}dx$ (multiple choice) 
Evaluate
  $\lim\limits_{h\to0}\int_{\frac{\pi}{4}}^{\frac{\pi}{4}+h}\frac{\sin(x)}{x}dx$
A. $0$
B. $\dfrac{\sqrt{2}}{2}$
C. $1$
D. $\dfrac{2\sqrt{2}}{\pi}$
E. $\dfrac{\sqrt{2}}{2\pi}$

I was under the assumption that I could plug in $0$ for $h$ and conclude that the integral is zero, making my answer choice A. However, the answer key states that D is the correct answer. 
I noticed that you could arrive at D by evaluating the integrand at $x=\dfrac{\pi}{4}$, but I'm not sure why that's justified. Is the key wrong, or is there a deeper layer of understanding to this question?
Any help is appreciated!
 A: $$L= \lim_{h \rightarrow 0} \int_{\pi/4}^{\pi/4+h} \frac{\sin x}{x} dx =0$$ As the lower anmd upper  limits of the integral coincide and $f(x)=\frac{\sin x}{x}$ is finte around (in the neighbour of) $x=\pi/4$. 
A: I agree with some of the comments on the question, that there must be a typo with a $1/h$ missing inside the limit. I will address two versions of the problem, and prove the answer rigorously in both cases.

Version I: The problem is correctly typeset, and it's just a trick question
If the question is indeed really meant as-is, then your answer of (A) is correct, and the key is wrong: the limit is $0$.
You can prove this easily by the Squeeze Theorem: think about a rectangle approximating the area under $\sin(x)/x$ between $\pi/4$ and $\pi/4+h$. Because $\sin(x)/x$ is decreasing at $x=\pi/4$, you can write the following bounds:
\begin{align}
  \frac{\sin(\pi/4+h)}{\pi/4+h}h \le \int_{\pi/4}^{\pi/4+h}\frac{\sin(x)}{x}\,dx \le \frac{\sin(\pi/4)}{\pi/4}h
\end{align}
Both the bounds $\to 0$ as $h\to 0$, so by the Squeeze Theorem, we have just proven that the integral converges to $0$ also.

Version II: There's a typo; someone forgot to typeset a $1/h$ in the limit
OTOH, I very strongly suspect a typo with a missing $1/h$, not only because this makes (D) the correct answer, but because that would also make this an excellent test question on the topic of the fundamental theorem of calculus...

Fundamental Theorem of Calculus:
Part (1): Let $f : [a, b] \to \mathbb{R}$ be continuous. Then,
$$\forall{x}\in [a, b]: \;\; 
  \biggl[\;\text{if}  \; {F(x) = \int_a^x f(t)\,dt}      \,,\;
         \;\text{then}\; {F'(x)=f(x)}                       \;   \biggr].$$
Part (2): Let $f : [a, b] \to \mathbb{R}$ be Riemann integrable. Then,
$$\forall{x}\in [a, b]: \;\; 
  \biggl[\;\text{if}  \; {F'(x)=f(x)}                    \,,\;
         \;\text{then}\; {\int_a^b f(x)\,dx = F(b)-F(a)}    \;   \biggr].$$

To summarize: parts (1) and (2) together say that antiderivative and the integral are one & the same. This means the derivative and the integral are essentially inverses of each other - if you take the derivative of the integral of $f$, they will ''cancel out'', and you'll just end up with $f$.
From here, you can probably see how the Fundamental Theorem applies to this question - let $f(x) = \sin(x)/x$. Then the $\lim_{h\to 0}$ strongly hints at a derivative, of an integral, of $f(x)\,dx$.
This version is also kind-of-sort-of a trick question -- in that it wants to trick you into looking for a closed-form expression for $\int{\sin(x)/x\,dx}$, even though this is not possible. That's why the Fundamental Theorem is key here -- you don't even need to know the closed form of the integral of $f$, because it's enough to know that the derivative you end up taking later will ultimately cancel it.
To solve the problem, let
$$
  A = \lim_{h\to 0}{ \dfrac{1}{h} \int_{\pi/4}^{\pi/4+h}f(x)\,dx}
$$
By the Fundamental Theorem, the integral of $f$ is equal to its antiderivative $F$, evaluated over the interval $\pi/4$ to $\pi/4+h$.
$$
  \int_{\pi/4}^{\pi/4+h}f(x)\,dx = F\left(\pi/4 + h\right) - F\left(\pi/4\right)
$$
Substitute this value of the definite integral into the expression for $A$, and obtain
$$
  A = \lim_{h\to 0}{\frac{F\left(\pi/4 + h\right) - F\left(\pi/4\right)}{h}}
$$
Finally, recognize that $A$ is literally just the definition of the derivative of $F(x)$ with respect to $x$ at $x=\pi/4$... and recall that $F$ was defined in the first place as a function such that $F'(x) = f(x)$. The very last step is, of course, the one where we need to know the form of $f(x)$ to evaluate $f\left(\pi/4\right)$.
$$
  A = F'\left(\frac{\pi}{4}\right) = f\left(\frac{\pi}{4}\right) 
    = \frac{\sin(\pi/4)}{\pi/4} = \boxed{\frac{2\sqrt{2}}{\pi}}
$$

A final footnote: notice how there are only two places in this answer where it even mattered that $f(x) = \sin(x)/x$ specifically:

*

*to establish that $f(x)$ was continuous at $\pi/4$, and

*to calculate the final answer in the very last step.

This is because the Fundamental Theorem of Calculus holds for all continuous $f$ - irrespective of which particular function $f$ might happen to be. That is why, to answer your original question, it's justified to evaluate the integrand at the (limit of the) limit, and get (D) as the answer.
