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}.$

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 $0$.

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