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I have a question from an exam written 18 years ago, and I can't solve it (I'm solving old questions to prepare for my own exam).

The Question

The question goes like so:

let g be a function, which has a derivative defined $\forall x\in [-1,1]$, and $g(0)=0,g'(0)=1$. Does the limit $$\lim_{x\rightarrow0}\frac{\sin\left(\int\limits _{x^{3}}^{x^{2}}\left(\int\limits _{0}^{t}g\left(s^{2}\right)ds\right)dt\right)}{x^{8}}$$ converge?

What have I tried?

I thought about using L'Hôpital's rule, and F.T.C. so I wrote:

$$\left(\int\limits _{x^{3}}^{x^{2}}\left(\int\limits _{0}^{t}g\left(s^{2}\right)ds\right)dt\right)'=2x\int\limits _{0}^{x^{2}}g\left(s^{2}\right)ds-3x^{2}\int\limits _{0}^{x^{3}}g\left(s^{2}\right)ds$$

but that didn't help much. I can even show that the limit is of the form $"\frac{0}{0}"$ to use the rule.

I tested this on $g(x)=x$ and it seems like this limit does converge (to 0 if I'm not mistaken)

I also thought Taylor was somehow related, but I really doubt it. I am also unsure of where the information we have about $g(0),g'(0)$ comes in.

BTW

this is what I got after applying L'Hôpital once

$$\lim_{x\rightarrow0}\frac{\left(2\int\limits _{0}^{x^{2}}g\left(s^{2}\right)ds-3x\int\limits _{0}^{x^{3}}g\left(s^{2}\right)ds\right)\cos\left(\int\limits _{x^{3}}^{x^{2}}\left(\int\limits _{0}^{t}g\left(s^{2}\right)ds\right)dt\right)}{8x^{6}}$$

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    $\begingroup$ You have only applied L'Hopital once. Why not keep using it? $\endgroup$ – user9464 Jan 19 at 20:39
  • $\begingroup$ That is just the numerator, L'Hopital has to be done on both. Consider the $x^8$ in the original numerator, and you will probably have to L'Hopital again. $\endgroup$ – Ninad Munshi Jan 19 at 20:39
  • $\begingroup$ @mrsamy I updated the answer and showed what I got after one iteration of L'Hopital, doing this 6 times more doesn't seem like a good option. Also, I don't even know how to show the limit is of form "0/0" (as I said), so Im not even sure if Im allowed to use the rule $\endgroup$ – snatchysquid Jan 19 at 20:45
  • $\begingroup$ @NinadMunshi please see the comment I wrote for mrsamy $\endgroup$ – snatchysquid Jan 19 at 20:45
  • $\begingroup$ It's incorrect, it should be a cosine off to the right $\endgroup$ – Ninad Munshi Jan 19 at 20:55
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This is an answer based on my comments.

Let $$G(t) =\int_{0}^{t}g(s^2)\,ds,F(u)=\int_{0}^{u}G(t)\,dt$$ Then we have $$F'(u) =G(u), G'(t) =g(t^2)$$ and hence by two applications of LHospital Rule we get $$\lim_{u\to 0}\frac{F(u)}{u^4}=\lim_{u\to 0}\frac{G(u)}{4u^3}=\lim_{u\to 0}\frac{g(u^2)}{12u^2}=\frac {1}{12}$$ The limit in question is $$\lim_{x\to 0}\frac{\sin(F(x^2)-F(x^3))}{x^8}$$ Since $F(x^2)\to 0,F(x^3)\to 0$ we can write the expression under limit as $$\frac {\sin(F(x^2)-F(x^3))}{F(x^2)-F(x^3)}\cdot \frac {F(x^2)-F(x^3)}{x^8}$$ and the first factor tends to $1$ so the limit equals the limit of second factor.

Now we can write $$\frac{F(x^2)-F(x^3)}{x^8}=\frac {F(x^2)}{x^8}-x^4\cdot\frac{F(x^3)}{x^{12}}$$ which tends to $$\frac {1}{12}-0\cdot\frac {1}{12}=\frac {1}{12}$$


LHospital Rule is a very powerful tool for evaluation of limits, but it has become infamous due to the crappy ways in which it is used frequently by beginners.

It is possible to avoid LHospital and use $\epsilon, \delta$ and integrating inequalities.

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  • $\begingroup$ Great explanation, thank you! the limit of $\frac{g(u^2)}{u^2}$ is obtained using L'hospitals rule, right? $\endgroup$ – snatchysquid Jan 19 at 22:17
  • $\begingroup$ @snatchysquid: no, that limit is by definition of derivative and $g'(0)=1$. $\endgroup$ – Paramanand Singh Jan 19 at 22:19
  • $\begingroup$ @snatchysquid: don't use LHospital more than necessary. $\endgroup$ – Paramanand Singh Jan 19 at 22:21
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Let us write

\begin{equation}g (x) = x h (x)\end{equation}

so that $h (x) \rightarrow 1$ when $x \rightarrow 0$. Let us first substitute $s = t u$ in the integral

\begin{equation}I \left(x\right) \colon = \int_{{x}^{3}}^{{x}^{2}}\left(\int_{0}^{t}g \left({s}^{2}\right) d s\right) d t = \int_{{x}^{3}}^{{x}^{2}}\left(\int_{0}^{t}{s}^{2} h \left({s}^{2}\right) d s\right) d t = \int_{{x}^{3}}^{{x}^{2}}{t}^{3} \left(\int_{0}^{1}u^2h \left({t}^{2} {u}^{2}\right) d u\right) d t\end{equation}

Now let us substitute $t = {x}^{2} v$, it follows that

\begin{equation}I \left(x\right) = {x}^{8} \int_{x}^{1}v^3\left(\int_{0}^{1} u^2h \left({x}^{4} {u}^{2} {v}^{2}\right) d u\right) d v \sim {x}^{8} \int_{0}^{1}{v}^{3} \left(\int_{0}^{1}u^2 d u\right) d v = \frac{{x}^{8}}{12}\end{equation}

As $\sin \left({\theta}\right) \sim {\theta}$ when ${\theta} \rightarrow 0$ it follows that the limit is $1/12$.

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    $\begingroup$ Shouldn't the limit be $1/12$? $\endgroup$ – Paramanand Singh Jan 19 at 21:50
  • $\begingroup$ @ParamanandSingh I'm not infallible. Is there a mistake in the above? $\endgroup$ – Gribouillis Jan 19 at 21:51
  • $\begingroup$ An easy way to check is using $g(x) =x$ $\endgroup$ – Paramanand Singh Jan 19 at 21:52
  • $\begingroup$ @ParamanandSingh I think the mistake is corrected. I forgot $u^2$ in the inner integral. Thanks. $\endgroup$ – Gribouillis Jan 19 at 21:58
  • $\begingroup$ Hello! Thanks for the answer, I have a few questions. what esnures us the existence of a function $h$ as described? In the last row, why did the integral bounds change from $x^2,x^3$ to $1,x$? And lastly, how and why are we allowed to used here the "~" symbol? $\endgroup$ – snatchysquid Jan 19 at 22:12

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