Differentiating definite integral seems weird to me. I've got this problem:
$$\lim_{k\to0}{\frac{1}{k}\int_0^k{(1+\sin{2x})^{\frac{1}{x}}dx}}$$
This is a Putnam problem from 1938. Link: https://mks.mff.cuni.cz/kalva/putnam/psoln/psol385.html
The solution suggests LH, because definite integral with the same bounds equals zero and k equals to zero as k tends to zero so this is 0/0.
Now the step should be:
$$\lim_{k\to0}\frac{(1+\sin{2k})^{\frac{1}{k}}}{1}$$
This can be further solved using logarithms and applying LH again. What i don't get is the step where we "canceled" the definite integral. 
If I integrated the function before applying LH, it must look like this:
$$\lim_{k\to0}\frac{F(k)-F(0)}{k}$$
without knowing what F (the primitive function to f) is. I can now apply LH: (let f denote the original function)
$$\lim_{k\to0}\frac{f(k)-f(0)}{1}$$
which is apparently different from what i should be getting.
I know that differentiating and integrating a function is an inverse process, but we can't apparently generally say, that:
$$\int_0^xf(t)dt=F(x)-F(0)$$
and
$$\int f(x)dx=F(x)$$
is the same.
What am i misunderstanding there?
 A: Use fundamental theorem of calculus. If the integral in question is denoted by $F(k) $ then the limit in question is $F'(0)$. The integrand $f(x) $ has a removable discontinuity at $x=0$ and lets redefine $f(0)=e^2$ to make it continuous at $0$. By FTC we have $F'(0)=f(0)=e^2$.
The use of L'Hospital's Rule is pretty roundabout here (since when did L'Hospital's Rule become a tool to compute derivatives?). I wonder why the solution manual suggests L'Hospital's Rule. 
A: Hint:
$$
\frac{d}{dk} \left( F(k)-F(0) \right) = f(k).
$$
A: That's the Fundamental Theorem of Calculus. It says that if $f$ is a real continuous function defined in the interval $[a,b]$ then the function $$F(x) = \int_a^x f(t)dt$$ is continuous in $[a,b]$, differentiable in $(a,b)$ and $$F'(x) = f(x).$$
In your example, we have $f(x) = (1+\sin(2x))^{\frac{1}{x}}$. This function isn't continuous at $x = 0$, but since $f(x) = e^{\ln f(x)}$, we can write $$f(x) = e^{\frac{\ln(1+\sin(2x))}{x}}$$
Using L'Hospital's rule, $$\lim_{x \rightarrow 0^+} \frac{\ln(1+\sin(2x))}{x} = \lim_{x \rightarrow 0^+} \frac{\frac{2\cos(2x)}{1+\sin(2x)}}{1} = 2,$$ so $\lim_{x \rightarrow 0^+} f(x) = e^2$ and the discontinuity at $0$ is actually removable. We can thus define $f(0) = e^2$, and the Fundamental Theorem allows us to differentiate $F(k)= \int_0^k f(x)dx$: $$F'(k) = f(k) =  (1+\sin(2k))^{\frac{1}{k}}$$
