Limit of one integral, how to use L'Hospital's rule I came across one problem, $$\lim_{x\to \overline{x}} \frac{\int_{x}^{\overline{x}}w(t)f(t)\,dt}{\int_{x}^{\overline{x}}w(t)g(t)\,dt}.$$ I know that $w(\overline{x})=0$ and $g(\overline{x})>0$ and $f(x)\neq 0$. If I use L'Hospital's rule, the limit would be $\frac{w(\overline{x})f(\overline{x})}{w(\overline{x})g(\overline{x})}$. But $w(\overline{x})=0$! So how to calculate this limit ? 
 A: Hint:
for $f(x)>0$ and $ g(\overline{x})\ne 0$
$$
\lim_{x \to \overline {x}}\frac{w(x)f(x)}{w(x)g(x)}=\lim_{x \to \overline {x}}\frac{f(x)}{g(x)} \qquad (1)
$$

$$\lim_{x\to \overline{x}} \frac{\int_{x}^{\overline{x}}w(t)f(t)\,dt}{\int_{x}^{\overline{x}}w(t)g(t)\,dt}=
\lim_{x\to \overline{x}} \frac{-\int^{x}_{\overline{x}}w(t)f(t)\,dt}{-\int^{x}_{\overline{x}}w(t)g(t)\,dt}
$$
Using L'Hopital and the fundamental theorem this is the limit $(1)$.
A: By the Fundamental Theorem of Calculus,
$$\frac{d}{dx} \int_x^{\bar x} h(t) \, dt = -h(x).$$
Combine this with L'Hôpital's rule.

EDIT
The limit is of the form $[\frac00]$, so L'Hôpital's rule can be used:
$$L := \lim_{x\to\bar x} \frac{\int_{x}^{\bar x} w(t) f(t) \, dt}{\int_{x}^{\bar x} w(t) g(t) \, dt}
= \lim_{x\to\bar x} \frac{\frac{d}{dx} \int_{x}^{\bar x} w(t) f(t) \, dt}{\frac{d}{dx} \int_{x}^{\bar x} w(t) g(t) \, dt}.$$
Now, by the Fundamental Theorem of Calculus,
$\frac{d}{dx} \int_x^{\bar x} h(t) \, dt = -h(x),$
so
$$L = \lim_{x\to\bar x} \frac{- w(x) f(x)}{- w(x) g(x)} = \lim_{x\to\bar x} \frac{f(x)}{g(x)}.$$
