6
$\begingroup$

Any hints on how to best approach this problem?

$$\lim_{x\to+\infty} \dfrac{1}{x} \int_{1}^{x} \dfrac{t^3}{1+t^3} dt$$

The first point of confusion for me is that $\dfrac{1}{x} \rightarrow 0$ as $ x \rightarrow +\infty$, so by evaluating the limit for $\dfrac{1}{x}$ and the integral separately and multiplying their limits afterwards should result in $0$, but I highly suspect that this is too simple of a solution that it must be wrong.

Secondly, my hunch is that to evaluate the limit of the integral I could find a function with a smaller area than $\dfrac{t^3}{1+t^3}$ on the listed interval and show that the limit tends to $\infty$ and this would be sufficient to show that $\dfrac{t^3}{1+t^3}$ must also tend to $\infty$ since it has a larger area. Is this the right approach and any hints as to how I could find a function with smaller area that I can show tends to $\infty$?

$\endgroup$
  • 1
    $\begingroup$ $\frac{t^3}{t^3+1}$ tends to $1$, which means that the integral of it tends to $\infty$ as the upper bound grows. $\endgroup$ – Arthur Jul 13 '17 at 15:52
  • 5
    $\begingroup$ Hint: L'Hospital's rule $\endgroup$ – user281392 Jul 13 '17 at 15:53
  • 5
    $\begingroup$ Write $t^3/(1+t^3)=1-1/(1+t^3)$ and use that the integral $\int_1^{+\infty}\frac{dt}{1+t^3}$ is convergent $\endgroup$ – Kelenner Jul 13 '17 at 15:55
  • 2
    $\begingroup$ One approach: $$\int_1^x\frac{t^3}{1+t^3}~\mathrm dt=\int_1^x\frac1{1+\frac1{t^3}}~\mathrm dt=\int_1^x1+\mathcal O(t^{-3})~\mathrm dt=x+\mathcal O(x^0)$$ And, $$\lim_{x\to\infty}\frac{x+\mathcal O(x^0)}x=1$$ $\endgroup$ – Simply Beautiful Art Jul 13 '17 at 15:56
16
$\begingroup$

$$\frac{1}{x}\int_{1}^{x}\frac{t^3}{1+t^3}\,dt = \frac{x-1}{x}-\frac{1}{x}\int_{1}^{x}\frac{dt}{1+t^3} = \color{red}{1}+O\left(\frac{1}{x}\right)\quad \text{as }x\to +\infty$$ since $f(t)=\frac{1}{1+t^3}$ is a positive function in $L^1(\mathbb{R}^+)$.
There is no need for de l'Hopital rule or subtler things, simple inequalities do the job just fine.

$\endgroup$
  • 1
    $\begingroup$ This is perhaps the simplest approach to the problem. +1 $\endgroup$ – Paramanand Singh Jul 13 '17 at 16:09
  • $\begingroup$ Very nice (+1). $\endgroup$ – hypergeometric Jul 13 '17 at 16:23
2
$\begingroup$

Note that since $$ \frac{t^3}{1+t^3} > \frac{1}{2} \quad \text{for } t > 1, $$ we clearly have $$ \int_{1}^{x} \dfrac{t^3}{1+t^3} dt \to \infty \quad \text{as } x \to \infty, $$ So you have $$ \lim_{x \to \infty} \frac{\int_{1}^{x} \frac{t^3}{1+t^3} dt}{x} $$ where top and bottom are both infinite and L'Hospital's Rule applies. Then use the Fundamental Theorem of Calculus for the top...

$\endgroup$
  • 2
    $\begingroup$ Your don't need to ensure that the integral tends to $\infty $ in order to use L'Hospital's Rule. Just denominator $x$ tending to $\infty$ is sufficient. $\endgroup$ – Paramanand Singh Jul 13 '17 at 16:08
  • $\begingroup$ @ParamanandSingh I don't think that's true. L'Hospital's Rule applies to indeterminate forms, either $0/0$ or $\infty/\infty$. If the integral converges, for example, the rule does not apply. $\endgroup$ – gt6989b Jul 13 '17 at 16:10
  • 2
    $\begingroup$ Like many you are not aware that the rule applies for $0/0$ and "$\text{anything} /\infty$" scenarios. Check Wikipedia article en.wikipedia.org/wiki/L%27H%C3%B4pital%27H rule and try to the proof given there. $\endgroup$ – Paramanand Singh Jul 13 '17 at 16:24
  • $\begingroup$ @gt6989b Paramanand is correct. LHR applies to $\frac{\text{anything}}{\infty}$, even if the limit of the "anything" fails to exist, provided the other conditions of LHR hold. $\endgroup$ – Mark Viola Jul 13 '17 at 17:11

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.