Find the limit of an integral with the help of the mean value theorem Find $$ \lim_{x \longrightarrow \pi} \frac{1}{x-\pi} \int^{\sin(x)}_{0} e^{t} \,dt$$
I know that the exponential function is continuous for all of the real numbers, so there exists c in the interval [0, sinx] with
$$ e^{c} = \frac{1}{\sin(x)} \int^{\sin(x)}_{0} e^{t} \,dt $$
But I dont know where to go from here. 
 A: You're almost there.  You have a number $c\in (0,\sin(x))$ and you have 
$$\int_0^{\sin(x)}e^t\,dt=e^c \sin(x)$$
Now divide by $x-\pi$ to arrive at
$$\frac{1}{x-\pi}\int_0^{\sin(x)}e^t\,dt=e^c \frac{\sin(x)}{x-\pi}$$
Finish by noting that $\sin(x)=-\sin(x-\pi)$, $\lim_{t\to 0}\frac{\sin(t)}{t}=1$, and $\lim_{x\to \pi}e^c=1$.

We now present two other methodologies.

METHODOLOGY $1$:
Good old L'Hospital's Rule is applicable here.  We have
$$\begin{align}
\lim_{x\to\pi}\frac{\int_0^{\sin(x)}e^t\,dt}{x-\pi}&=\lim_{x\to\pi}\left(\cos(x)e^{\sin(x)}\right)\\\\
&=-1
\end{align}$$

METHODOLOGY $2$:
Alternatively, note that 
$$\begin{align}
\frac1{x-\pi}\int_0^{\sin(x)}e^t\,dt&=\frac{e^{\sin(x)}-1}{x-\pi}\\\\
&=\frac{e^{-\sin(x-\pi)}-1}{x-\pi}\\\\
&=\left(\frac{e^{-\sin(x-\pi)}-1}{\sin(x-\pi)}\right)\left(\frac{\sin(x-\pi)}{x-\pi}\right)\\\\
\end{align}$$
Then, use the "standard" limits $\lim_{t\to0}\frac{e^{-t}-1}{t}=-1$ and $\lim_{x\to0}\frac{\sin(t)}{t}=1$
A: Let $$ F(x)=\int_0^{\sin x}e^t dt $$
then $F(\pi)=0$ and $F$ is derivable on $[0,\pi]$. Thus
$$ \frac{1}{x-\pi}\int_0^{\sin x}e^t dt=\frac{F(x)-F(\pi)}{x-\pi}\underset{x\rightarrow\pi}{\longrightarrow}F'(\pi)=\cos(\pi)e^{\sin(\pi)}=-1 $$
You can also say that
$$ \frac{1}{\sin x}\int_0^{\sin x}e^t dt=e^{c_x} $$
with $c_x\in[0,\sin x]$. By the squeeze theorem you have $\lim\limits_{x\rightarrow\pi}c_x=0$ and thus
$$ \frac{F(x)}{x-\pi}=\frac{\sin(x)-\sin(\pi)}{x-\pi}e^{c_x}\underset{x\rightarrow\pi}{\longrightarrow}\sin'(\pi)e^{0}=-1 $$ 
A: MVT for integrals
$\displaystyle{\int_{0}^{\sin x}}e^tdt=e^s\int_{0}^{\sin x}1dt=$
$e^s (\sin x-0)$; where $s \in [0,\sin x]$.
Note: $\lim_{x \rightarrow π} s =0$;
$\lim_{x \rightarrow π}e^s \dfrac{1}{x-π} \sin (π-x) =$
$(-1)\lim_{x \rightarrow π}e^s \dfrac {\sin (x-π)}{x-π}=$
$(-1)\lim_{x\rightarrow π}e^s \cdot \lim_{x \rightarrow π}\dfrac{\sin (x-π)}{x-π}.$
Can you finish?
