1
$\begingroup$

Evaluate the limit $$\lim\limits_{x \to 0} \frac{\displaystyle{\int\limits_0^{x^2 } {\frac{{\mathrm dt}}{{\sqrt[5]{{t^2 + 6}}}}} }}{{x^2 }}.$$

I have a real big limit problem here. I have no idea how to solve it. This one is like a hybrid between Limit and Integral (sorry for my English). There are also two variables, $x$, and $t$. Can you explain the method used and the steps? Thanks in advance.

$\endgroup$
4
  • $\begingroup$ Do you know L'Hôpital's rule? $\endgroup$
    – Gary
    Apr 26 at 12:57
  • $\begingroup$ yeah, i do know it $\endgroup$
    – too-much
    Apr 26 at 12:57
  • 1
    $\begingroup$ Ok, then by the fundamental theorem of calculus and the chain rule, $\frac{d}{{dx}}\int_0^{x^2 } {f(t)dt} = 2xf(x^2 )$. Can you finish? $\endgroup$
    – Gary
    Apr 26 at 13:00
  • $\begingroup$ Thank you, i can finish this problem. $\endgroup$
    – too-much
    Apr 26 at 13:14

3 Answers 3

5
$\begingroup$

Do not use l'Hospital's rule, it does not apply here, it is a circular argument. The expression is just a difference quotient $$ \frac{F(x^2)-F(0)}{x^2-0} $$ for a primitite $F$ with $F'(t)=\frac1{\sqrt[5]{6+t^2}}$.

The limit of the difference quotient is already the differential quotient or derivative $F'(0)=\frac1{\sqrt[5]6}$, there is no need to involve circular arguments.


Or put another way, by the integral mean value theorem $$ \int_0^{x^2}\frac1{\sqrt[5]{6+t^2}}\,dt=(x^2-0)\frac1{\sqrt[5]{6+c^2}} $$ for some $0<c<x^2$. In the limit, $c$ converges to $0$.

$\endgroup$
0
$\begingroup$

Hint: Since it resembles the indeterminate form $0/0$, use L'Hopital's rule, and use Differentiation under the Integral Sign to differentiate the integral expression in the numerator.

$\endgroup$
0
$\begingroup$

You can use L'Hôpital's rule, or you can just note that the integrand $\dfrac{1}{\sqrt[5]{t^2+6}}$ tends to the constant $\dfrac{1}{\sqrt[5] 6}$ as $t\to 0$. So the integral is equal to $\dfrac{x^2}{\sqrt[5] 6}+o(x^2)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.