Question containing limit,derivative and definite integration

Question

Problem: Let $f: \Bbb R \to \Bbb R $ be a differential function such that $f(3)=3$ and $f'(3)= \frac {1}{2}$

Then what is the value of $\lim _{x \to 3}$ $ \int^{f(x)}_{3} \frac{2 t^3}{x-3} dt $

Solution:

$\lim _{x \to 3}$ $ \int^{f(x)}_{3} \frac{2 t^3}{x-3} dt $

$\lim _{x \to 3} \frac{2 }{x-3} \int^{f(x)}_{3} t^3 dt $

value of f(x) can be calculated using $f(3)=3$ and $f'(3)= \frac {1}{2}$

I tried first principle method of differential but didn't get it

Answer 1

OK. You've gotten to $$ \lim _{x \to 3} \frac{2 }{x-3} \int^{f(x)}_{3} t^3 dt $$

Now look at the integral. The antiderivative of $t^3$ is just $\dfrac{t^4}{4}$. So your expression becomes $$ \lim _{x \to 3} \frac{2 }{x-3} \left( \dfrac{f(x)^4 - 3^4}{4} \right) $$

In that expression, both numerator and denominator approach zero. Can you think of a rule to help you compute the limit of the ratio?

Answer 2

You can write your limit as $$ \lim _{x \to 3}2 \frac{\displaystyle\int^{f(x)}_{3} t^3\, dt}{x-3} $$ which is in the form $0/0$, because by hypothesis $f$ is continuous at $3$. Then you can apply l'Hôpital's theorem: $$ \lim _{x \to 3}2 \frac{\displaystyle\int^{f(x)}_{3} t^3\, dt}{x-3} \overset{\mathrm{H}}{=} \dots $$ If $F(x)=\int_{3}^{x}t^3\,dt$, you know that $F'(x)=x^3$, by the fundamental theorem of calculus. Just realize that the numerator is $$ F(f(x)) $$ and use the chain rule.