Evaluate $ \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}.$ Problem
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}.$
Solution
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?
 A: 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 $0$.
A: 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 $1$.
A: 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
$$\lim_{x \to 0}\frac{f(x)}{x}= \lim_{x \to 0}\frac{f(x)-f(0)}{x-0}=f'(0)$$
A: 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 $$\lim\limits_{x \to 0}\frac{\int_0^{x^2}f(t){\rm d}t}{x^2\int_0^x f(t){\rm d}t}=1$$
A: 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 $0$.
