Let $ f $ be a function defined on a real interval from $0 $ to $1$ and have a discontinuity at $1/2$ (however the righthand and lefthand limits still exist). Let $ F $ be a function defined by $ \int^x_0 fdt$. Then compute the righthand limit of $(F(x)-F(1/2))/(x-1/2$) as $ x $ approaches $1/2$ from the right side.

I want to use the following theorem (L'Hospital's Rule) but can't come up with an argument.

Let $ f $ and $g$ be differentiable in $(a,b)$ and $g'(x)$ be nonzero for all of $x\in(a,b)$. Suppose $f'(x)/g'(x) \rightarrow A$ as $ x \rightarrow a $. If $f(x)\rightarrow 0$ and $g(x)\rightarrow 0$ as $x \rightarrow a$ or if $g(x) \rightarrow \infty$ as $x \rightarrow a$ then $f(x)/g(x) \rightarrow A$ as $x \rightarrow a$.

  • $\begingroup$ since $F(x)-F(1/2)=\int_{1/2}^{x}f \,dt$, the discontinuity at $1/2$ doesn't matter. You can use the fundamental theorem of calculus. $\endgroup$ – Bananach Mar 8 '15 at 18:08
  • $\begingroup$ True, but I'd like to use L'Hospital's rule $\endgroup$ – Fry Mar 8 '15 at 18:12
  • $\begingroup$ L'Hospital's rule asks you to compute a derivative --- of the integral. I don't think there is a way to avoid that $\endgroup$ – Bananach Mar 8 '15 at 18:20
  • $\begingroup$ yes that's what confused me.. is it possible to do that? $\endgroup$ – Fry Mar 8 '15 at 18:22
  • $\begingroup$ I would still be using the fundamental thm of calculus but I can't figure out how to use L'Hospital's rule.I'm not sure what I'm missing $\endgroup$ – Fry Mar 8 '15 at 18:48

First note that $$ \frac{d}{dx}\int_a^x f(t)\ dt = \frac{d}{dx}\left[F(x)-F(a)\right] $$ $$= \frac{d}{dx}F(x) - \frac{d}{dx}F(a) = f(x) - 0 = f(x) $$ And $$ \frac{d}{dx}\int_a^b f(t)\ dt = \frac{d}{dx}\left[F(b)-F(a)\right] $$ $$= \frac{d}{dx}F(b) - \frac{d}{dx}F(a) = 0 - 0 = 0$$ Therefore $$ \lim\limits_{x\to\frac12^+} \left[\frac{F(x)-F\left(\frac12\right)}{x -\frac12}\right]$$ $$= \lim\limits_{x\to\frac12^+} \left[\frac{\frac{d}{dx}\left[F(x)-F\left(\frac12\right)\right]}{\frac{d}{dx}\left[x -\frac12\right]}\right]$$ $$= \lim\limits_{x\to\frac12^+} \left[\frac{\frac{d}{dx}F(x)-\frac{d}{dx}F\left(\frac12\right)}{\frac{d}{dx}x -\frac{d}{dx}\frac12}\right]$$ $$=\lim\limits_{x\to\frac12^+} \left[\frac{f(x)-0}{1 -0}\right]=\lim\limits_{x\to\frac12^+} f(x)$$ Note that $F\left(\frac12\right)$ produces a constant and the derivative of a constant is zero.

  • 1
    $\begingroup$ nice explanation! $\endgroup$ – user42 Mar 9 '15 at 17:26

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