Limit with integral - L'Hospital's rule Find 
$$ \lim_{x \rightarrow 0^+} \frac{\int_{0}^{x^2} (1 + \sin(t))^{1/t} dt}{x \sin(x)} $$
Let 
$$ F(t) \mbox{ such that } F'(t) = (1 + \sin(t))^{1/t}  $$
Then we use L'Hospital's rule:
$$ \lim_{x \rightarrow 0^+} \frac{\int_{0}^{x^2} (1 + \sin(t))^{1/t} dt}{x \sin(x)} = \frac{F'(x^2) - F'(0) }{\sin(x) + x \cos(x)} $$
but $$ F'(0) $$ is not defined (we have $1/t$ part as exponent)
 A: Lebnitz rule is applied as follows: 
$$\frac{d}{dx} \int_{f(x)}^{g(x)} h(t) dt$$
$$=h(f(x))\times f'(x) - h(g(x)) \times g'(x) \quad (1)$$
Now w.r.t to your question, 
$$f(x) = 0$$
$$g(x) = x^2$$
$$h(t) = (1+\sin(t))^{1/t}$$
Now we'll prove that the following limit exists: 
$$L = \lim_{t \to 0} h(t)$$
$$L = \lim_{t \to 0} (1+\sin(t))^{1/t}$$
For this we'll first evaluate the limit for $\log h(t)$. 
$$\lim_{t \to 0} \log(h(t))$$
$$=\lim_{t \to 0}\frac{\log(1+\sin(t))}{t}$$
Now upon applying L'Hopital rule to the above limit we can establish that $L=e$
Therefore, 
$$\lim_{t \to 0} (1+\sin(t))^{1/t} = e$$
Now substituting this back to equation (1), using $f$, $g$, $h$, mentioned above, we get: 
$$\frac{d}{dx} \int_{0}^{x^2} (1+\sin(t))^{1/t}$$
$$=((1+\sin(x^2))^{1/x^2})\times 2x - (\lim_{t\to 0}(1+\sin(t))^{1/t})\times 0$$
$$=((1+\sin(x^2))^{1/x^2})\times 2x$$
Hope this helps.
A: Hint:
$$\lim_{x \to 0^+}\frac{\int_0^{x^2} (1+\sin(t))^{1/t} \,dt}{x\sin(x)}=\lim_{x \to 0^+}\frac{2x (1+\sin(x^2))^{x^{-2}}}{\sin(x) + x\cos(x)}=2\lim_{x \to 0^+}\frac{(1+\sin(x^2))^{x^{-2}}}{\frac{\sin(x)}{x}+\cos(x)}.$$
A: If the integrand is denoted by $f(t) $ then $\lim_{t\to 0^{+}}f(t)=e$ (check this). If we define $f(0)=e$ then $f$ becomes continuous at $0$.
Next the expression under limit can be written as $$\frac{x}{\sin x} \cdot\frac{1}{x^2}\int_{0}^{x^2}f(t)\,dt$$ The first factor tends to $1$ and thus the desired limit is equal to $$\lim_{u\to 0^+}\frac{1}{u}\int_{0}^{u}f(t)\,dt$$ via substitution $u=x^2$. Since $f$ is continuous at $0$ by fundamental theorem of calculus the above limit is $f(0)=e$.
