Let $f(x)$ be continuously differentiable. $f(0)=0$ and $f'(0) \neq 0$. Evaluate $ \lim\limits_{x \to 0}\dfrac{\int_0^{x^2}f(t){\rm d}t}{x^2\int_0^x f(t){\rm d}t}.$


Consider applying l'Hôpital's rule.

\begin{align*} \lim_{x \to 0}\frac{\int_0^{x^2}f(t){\rm d}t}{x^2\int_0^x f(t){\rm d}t}&=\lim_{x \to 0}\frac{2xf(x^2)}{2x\int_0^x f(t){\rm d}t+x^2f(x)}\\ &=2\lim_{x \to 0}\frac{f(x^2)}{2\int_0^x f(t){\rm d}t+xf(x)}\\ &=2\lim_{x \to 0}\frac{2xf'(x^2)}{2f(x)+f(x)+xf'(x)}\\ &=4\lim_{x \to 0}\frac{xf'(x^2)}{3f(x)+xf'(x)}\\ \end{align*}

It seems to be impossible to go on from here. Is it a wrong question?

  • 1
    $\begingroup$ Try L'Hospital's one more time? $\endgroup$ – T.J. Gaffney Jan 12 '19 at 3:33
  • $\begingroup$ @Gaffney you are not given that $f(x)$ is 2-order differentiable. How could you try L'Hospital from the last line? $\endgroup$ – mengdie1982 Jan 12 '19 at 3:37
  • 1
    $\begingroup$ Then how about the definition of $f'(0) $? $\endgroup$ – xbh Jan 12 '19 at 3:45

Continued from where you got stuck: $$ \cdots = 4\lim_{x\to 0}\frac {f'(x^2)} {3\dfrac {f(x)}x +f'(x) }= \frac {4f'(0)}{4f'(0)} = 1, $$

where $$ \lim_{x\to 0} \frac {f(x)}x = \lim_{x \to 0} \frac {f(x) -f(0)}{x - 0}, $$ and $f'$ is continuous at $0$.

  • $\begingroup$ Thanks. I forget that! $\endgroup$ – mengdie1982 Jan 12 '19 at 4:05

We have

$$ \lim_{x\to 0} \frac{\int_{0}^{x^2}f(t) \, \mathrm{d}t}{x^2 \int_{0}^{x} f(t) \, \mathrm{d}t} = \lim_{x\to 0} \frac{\int_{0}^{x^2} f(t) \, \mathrm{d}t / x^4}{\int_{0}^{x} f(t) \, \mathrm{d}t / x^2}. \tag{*} $$

Now by the L'Hospital's rule,

  • $\displaystyle \lim_{x \to 0} \frac{\int_{0}^{x^2} f(t) \, \mathrm{d}t}{x^4} = \lim_{x \to 0} \frac{2x f(x^2)}{4x^3} = \frac{f'(0)}{2} $,

  • $\displaystyle \lim_{x \to 0} \frac{\int_{0}^{x} f(t) \, \mathrm{d}t}{x^2} = \lim_{x \to 0} \frac{f(x)}{2x} = \frac{f'(0)}{2} $.

Therefore both the denominator and the numerator of $\text{(*)}$ converge to the same non-zero value and hence the answer is $1$.


Hint Divide both top and bottom by $x$:

$$4\lim_{x \to 0}\frac{xf'(x^2)}{3f(x)+xf'(x)}=4\lim_{x \to 0}\frac{f'(x^2)}{3\frac{f(x)}{x}+f'(x)}$$

Note that since $f'$ is continuous you have $$\lim_{x \to 0}f'(x^2)=\lim_{x \to 0}f'(x)=f'(0)$$

Also $$\lim_{x \to 0}\frac{f(x)}{x}= \lim_{x \to 0}\frac{f(x)-f(0)}{x-0}=f'(0)$$

  • $\begingroup$ $f(x)$ is continuously differentiable, but the same is not necessarily true for $f'(x)$. $\endgroup$ – JavaMan Jan 12 '19 at 3:49
  • $\begingroup$ @JavaMan Continuously differentiable means that $f'$ is continuous... I only used that $f'$ is continuous at $x=0$. I never used that $f'$ is continuously differentiable. $\endgroup$ – N. S. Jan 12 '19 at 3:51
  • $\begingroup$ Your answer was a good one, but your answer explicitly states that “since $f’(x)$ is continuously differentable...” which is false. I understand that you both didn’t need and didn’t use the fact that $f \in C^1$. $\endgroup$ – JavaMan Jan 13 '19 at 3:16
  • $\begingroup$ @JavaMan typo, fixed. ty $\endgroup$ – N. S. Jan 13 '19 at 3:21
  • $\begingroup$ No problem, and I wasn’t trying to put down your answer. I’d upvote it but I’d already done that! $\endgroup$ – JavaMan Jan 13 '19 at 3:21

Consider $$F(x)=\int_0^xf(t)dt$$ Then $F$ has the following Taylor expansion: For all $x$ in the neighborhood of $0$, there exists $c_x$ such that $|c_x|\leq |x|$ and $$F(x)=F(0)+xF^\prime(0)+\frac{x^2}2 F^{\prime\prime}(c_x)=\frac{x^2}2f^\prime(c_x)$$ Therefore $$\begin{split} \dfrac{\int_0^{x^2}f(t){\rm d}t}{x^2\int_0^x f(t){\rm d}t} &= \frac{F(x^2)}{x^2F(x)}\\ &= \frac{x^4f^\prime(c_{x^2})}{x^4f^\prime(c_x)}\\ &= \frac{f^\prime(c_{x^2})}{f^\prime(c_x)}\\ \end{split}$$ And since $f^\prime$ is continuous, both numerator and denominator tend to $f^\prime(0)$ as $x\rightarrow 0$.

Thus $$\lim\limits_{x \to 0}\frac{\int_0^{x^2}f(t){\rm d}t}{x^2\int_0^x f(t){\rm d}t}=1$$


Via definition of derivative we have $$f(t) =f(0)+tf'(0)+o(t)=tf'(0)+o(t)$$ and integrating the above (this involves the use of L'Hospital's Rule) we get $$\int_{0}^{x}f(t)\,dt=\frac{x^2}{2}f'(0)+o(x^2)\tag{1}$$ and thus $$\int_{0}^{x^2}f(t)\,dt=\frac{x^4}{2}f'(0)+o(x^4)\tag{2}$$ By the equations $(1),(2)$ we can see that the desired limit is $1$.

Note: The above assumes that $f$ is continuous in some neighborhood of $0$ and differentiable at $0$.


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