Exercise of fundamental theorem of calculus

Question

Let $f$ is continuous on $[0,1]$, then

$\lim\limits_{x\rightarrow0 ^+} x^2\int_{x}^{1} \frac{f(t)}{t^3} dt~ = ~? $

The problem is form chapter of FTC, but I have no idea of this problem. And i think that the solution is depend on the function $f$ ?

Answer 1

Use L'Hôpital's rule (and FTC): $$\lim\limits_{x\rightarrow0 ^+} x^2\int_{x}^{1} \frac{f(t)}{t^3} dt= \lim\limits_{x\rightarrow0 ^+} \frac{\int_{x}^{1} \frac{f(t)}{t^3} dt}{1/x^2}\overset{\text{L'Hôpital}}{=} \lim\limits_{x\rightarrow0 ^+} \frac{-\frac{f(x)}{x^3}}{-2x^{-3}}=\frac{f(0)}{2}.$$

P.S. It is not necessary to check that numerator goes to infinity. See How does the L.Hopital rule work when numerator $\neq$ $\infty$. Anyway, you can also say that if numerator does not go to infinity then necessarily $f(0)=0$ (otherwise $\frac{f(t)}{t^3}\sim \frac{f(0)}{t^3}$ whose improper integral in a right neighborhood of $0$ is not convergent) and the limit is easily seen to be $0$. So the result $f(0)/2$ holds also in this case.

Answer 2

Here is another take on this without the use of L'Hospital's Rule. We have \begin{align} g(x) &= x^{2}\int_{x}^{1}\frac{f(t)}{t^{3}}\,dt\notag\\ &= x^{2}\int_{x}^{1}\frac{f(t) - f(0)}{t^{3}}\,dt + x^{2}f(0)\int_{x}^{1}\frac{dt}{t^{3}}\notag\\ &= x^{2}\int_{x}^{1}\frac{f(t) - f(0)}{t^{3}}\,dt + x^{2}f(0)\left(\frac{1}{2x^{2}} - \frac{1}{2}\right)\notag\\ &= \frac{f(0)}{2} - \frac{x^{2}f(0)}{2} + x^{2}\int_{x}^{1}\frac{f(t) - f(0)}{t^{3}}\,dt\notag \end{align} As $x \to 0^{+}$ we can see that the first term above remains constant, the second term goes to $0$ and we shall prove below that the third term also goes to $0$ and hence the desired limit is $f(0)/2$.

Let $\epsilon > 0$ be given. Then by continuity of $f$ at $0$ it follows that there is an $h > 0$ such that $$|f(t) - f(0)| < \epsilon$$ for all $t$ with $0 \leq t \leq h$ and hence if $0 < x < h$ then \begin{align} h(x) &= \left|x^{2}\int_{x}^{1}\frac{f(t) - f(0)}{t^{3}}\,dt\right|\notag\\ &\leq \left|x^{2}\int_{x}^{h}\frac{f(t) - f(0)}{t^{3}}\,dt\right| + \left|x^{2}\int_{h}^{1}\frac{f(t) - f(0)}{t^{3}}\,dt\right|\notag\\ &< \epsilon x^{2}\int_{x}^{h}\frac{dt}{t^{3}} + x^{2}M(h)\notag\\ &= \epsilon x^{2}\left(\frac{1}{2x^{2}} - \frac{1}{2h^{2}}\right) + x^{2}M(h)\notag\\ &< \frac{\epsilon}{2} + x^{2}M(h)\notag \end{align} where $$M(h) = \left|\int_{h}^{1}\frac{f(t) - f(0)}{t^{3}}\,dt\right|$$ is independent of $x$. By the above equation it is clear that $$\limsup_{x \to 0^{+}}h(x) \leq \frac{\epsilon}{2}$$ Since $\epsilon > 0$ was arbitrary it follows that $\lim_{x \to 0^{+}}h(x) = 0$ and we are done.