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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 ?

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  • $\begingroup$ There are many missing hypotheses. Please include them. $\endgroup$ – zhw. Jun 10 '17 at 17:22
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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 $(1)$.

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

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

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