# 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 ?

• There are many missing hypotheses. Please include them. – zhw. Jun 10 '17 at 17:22

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 $(1)$.

• Are you suggesting that the original limit problem is equal to $\lim_{x \to \overline{x}} \frac{w(x)f(x)}{w(x)g(x)}=\lim_{x \to \overline{x}} \frac{f(x)}{g(x)}$ ? – congmingniao Jun 10 '17 at 13:50
• I added to my answer. I hope it's useful :) – Emilio Novati Jun 10 '17 at 13:56
• Thank you. Besides $w(\overline{x})=0$, $w(x)=0$ for some $x<\overline{x}$. can I still cancel out $w(x)$ in $\lim_{x \to \overline{x}}\frac{-w(x)f(x)}{-w(x)g(x)}$? – congmingniao Jun 10 '17 at 14:03
• Yes you can. It is similar to $lim_{x \to 1}\frac{x^2-1}{x-1}$ – Emilio Novati Jun 10 '17 at 14:06
• Is $\lim_{x \to \overline{x}} \frac{f(x)}{g(x)}=\frac{f(\overline{x})}{g(\overline{x})}$ right when f(x) and g(x) are continuous? – congmingniao Jun 10 '17 at 14:29

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 $$L = \lim_{x\to\bar x} \frac{- w(x) f(x)}{- w(x) g(x)} = \lim_{x\to\bar x} \frac{f(x)}{g(x)}.$$

• could you please give more details ? I cannot catch your idea. Thank you. – congmingniao Jun 10 '17 at 13:48
• @congmingniao: The post have been edited. – md2perpe Jun 10 '17 at 13:56
• Thank you. Besides $w(\overline{x})=0$, $w(x)=0$ for some $x<\overline{x}$. can I still cancel out $w(x)$ in $\lim_{x \to \overline{x}}\frac{-w(x)f(x)}{-w(x)g(x)}$ ? – congmingniao Jun 10 '17 at 14:02
• As long as you take $x \neq \bar x$ during taking the limit. – md2perpe Jun 10 '17 at 14:12
• Is $\lim_{x \to \overline{x}} \frac{f(x)}{g(x)}=\frac{f(\overline{x})}{g(\overline{x})}$ right when f(x) and g(x) are continuous? – congmingniao Jun 10 '17 at 14:31