If $a,b,c$ are real numbers then the value of the integral If $a,b,c$ are real numbers then compute the value of 
$$
\lim_{t\to0}\ln\left(\frac{1}{t}\int_0^t(1+a\sin(bx))^{c/x}dx\right)
$$
My Thoughts:
Tried some substitutions. Like $1/x=t$.Doesn't help.There was a similarity to Riemann sums too. But it didn't turn out to be that.
Any hints/suggestions from your side?
 A: This looks like the Fundamental Theorem of Calculus in disguise. If we let $f(x)=(1+a\sin(bx))^{c/x}$ and $F(x)$ equal the antiderivative of $f(x)$ and we move the limit inside the logarithm, we have $$\ln\left({\lim_{t\to0}\frac{F(t)-F(0)}{t-0}}\right)$$ which equals the natural logarithm of the derivative of $F(t)$ at $t=0$, which is just $\lim_{x\to0}\ln(f(x))$. This leaves us with
$$\lim_{x\to0}\ln((1+a\sin(b*x))^{c/x})=\lim_{x\to0}\frac{c*\ln(1+a\sin(b*x))}{x}$$
From there, you should be able to use L'Hopital's rule to force an answer out of this question.
A: The integrand function is not defined at $0$, so before starting to compute the limit, we have to ensure the integral exists.
Since later we're interested in the the limit at $0$, we can just work in some unspecified neighborhood of $0$. In particular, we take it so that $1+a\sin(bx)>0$, which is possible because $\sin0=0$ and the sine is continuous.
If we prove that
$$
\lim_{x\to0}(1+a\sin(bx))^{c/x}
$$
is finite, then we can extend the function by continuity and the integral will pose no problem of existence. Such a limit can be computed by first taking the logarithm, so let's study
$$
\lim_{x\to0}\ln\bigl((1+a\sin(bx))^{c/x}\bigr)=
\lim_{x\to0}\frac{c\ln(1+a\sin(bx))}{x}
$$
Well, this is the derivative at $0$ of the function $g(x)=c\ln(1+a\sin(bx))$; since
$$
g'(x)=\frac{abc\cos(bx)}{1+a\sin(bx)}
$$
and $g'(0)=abc$, we have that
$$
\lim_{x\to0}(1+a\sin(bx))^{c/x}=e^{abc}
$$
so our problem becomes to find
$$
\lim_{t\to0}\ln\left(\frac{1}{t}\int_0^t f(x)\,dx\right)
$$
where (for $x$ in the above mentioned neighborhood of $0$)
$$
f(x)=\begin{cases}
(1+a\sin(bx))^{c/x} & \text{if $x\ne0$}\\[4px]
e^{abc} & \text{if $x=0$}
\end{cases}
$$
which is continuous.
Now apply continuity of $\ln$ and the fundamental theorem of calculus.
