$\lim_{h \to 0} \frac{1}{h} \int _{\sin x} ^{\sin (x+h)} e^{\cos(x+h)t}\ dt$ Find the limit value of
$$\lim_{h \to 0} \frac{1}{h} \int _{\sin x} ^{\sin (x+h)} e^{\cos(x+h)t}\ dt$$
 A: For some $c$ lying between $\sin x$ and $\sin (x+h)$, 
$$
\int_{\sin x}^{\sin (x+h)} e^{\cos (x+h)t}\, dt = \left( \sin (x+h)-\sin x \right) e^{\cos (x+h)c}.
$$
Dividing by $h$ and taking limits yields
$$
\cos x \ e^{\cos x \sin x}.
$$
A: For the antiderivative $$ \int e^{\cos(x+h)t}\ dt=\sec (h+x) e^{t \cos (h+x)}$$ So, for the integral $$\int_{\sin x}^{\sin (x+h)} e^{\cos (x+h)t}\, dt = \sec (h+x) \left(e^{\sin (h+x) \cos (h+x)}-e^{\sin (x) \cos (h+x)}\right)$$ for which the Taylor expansion built at $h=0$ is $$h \cos (x) e^{\sin (x) \cos (x)}+\frac{1}{8} h^2 e^{\sin (x) \cos (x)} (-4 \sin
   (x)+\cos (x)+3 \cos (3 x))+O\left(h^3\right)$$ and then the result given by Siminore.
A: Another approach is to use L' Hopital's Rule in combination with Leibniz Integral Rule.
The limit can be treated as a $\displaystyle \frac{0}{0}$ indeterminate form.
Let $\displaystyle f(x,h,t) = e^{\cos(x+h)t}$ and $\displaystyle I(x,h) = \int_{\sin x}^{\sin(x+h)}f(x,h,t)dt$
By L' Hopital's Rule, the limit is equal to $\displaystyle \lim_{h \to 0} \frac{\frac{dI}{dh}}{\frac{dh}{dh}} = \lim_{h \to 0} \frac{dI}{dh}$
By Leibniz Rule, $\displaystyle \frac{dI}{dh} = \frac{d}{dh}(\int _{\sin x} ^{\sin (x+h)} e^{\cos(x+h)t}\ dt)\\=\displaystyle  \int _{\sin x} ^{\sin (x+h)}\frac{\partial{f}}{\partial{h}}dt + f(x, h,\sin(x+h))\frac{d\sin(x+h)}{dh} - f(x, h, \sin x)\frac{d\sin x}{dh}$
It can be shown that the first and last terms vanish as $h \to 0$ giving the limit as $\displaystyle \lim_{h \to 0} \cos(x+h)e^{\cos(x+h)\sin(x+h)} = \cos xe^{\cos x\sin x}$
