We need to evaluate the following limit:

$$\lim\limits_{x\to 0} \frac{1}{x} \int_0^x \sin^2\left(\frac{1}{y}\right)\,\mathrm dy.$$

I am finding hard to solve this integral as I cannot see a clear method to evaluate such integral. I have tried using substitutions, integration by parts, expansion formulas etc. but I don't see them be of any help. Also, I cannot take the limit inside. The best I could do was to reduce it to cosine function as follows:

$$\lim\limits_{x\to 0}\frac{1}{x} {\int}_0^x \sin^2\left(\frac{1}{y}\right)\,\mathrm dy = \lim\limits_{x\to 0}\frac{1}{x} {\int}_0^x \frac{\left(1-\cos\left(\frac{2}{y}\right)\right)}{2}\,\mathrm dy = \frac{1}{2}-\lim\limits_{x\to 0}\frac{1}{x} {\int}_0^x \cos\left(\frac{2}{y}\right)\,\mathrm dy.$$

But again the integrand has no standard antiderivative. Is there some easy method to solve this or I am making some mistakes? Any hint/help will be of great help.

Thanks in advance.

  • 2
    $\begingroup$ Try to use L'hospitals rule. $\endgroup$ – Mustafa Said Apr 23 at 16:43
  • $\begingroup$ That will yield that limit is not defined? $\endgroup$ – Rick Apr 23 at 16:51
  • $\begingroup$ I am so sorry, in the limit x goes to 0. $\endgroup$ – Rick Apr 23 at 16:54
  • 1
    $\begingroup$ Your approach is correct and the last term tends to $0$ so that the desired limit is $1/2$. Proving that the last term $(1/x)\int_{0}^{x}\cos(2/y)\,dy$ tends to $0$ is tricky. You need to analyze the function $F(x) =x^2\sin(2/x),F(0)=0$ which gives $F'(0)=0,F'(x)=2x\sin(2/x)-2\cos(2/x)$. $\endgroup$ – Paramanand Singh Apr 25 at 4:37
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    $\begingroup$ Ok consider function $G(x) =\int_{0}^{x}\cos(2/y)\,dy$ then $G'(0)=\lim_{x\to 0}(1/x)\int_{0}^{x}\cos(2/y)\,dy$. From last comment and FTC we get $F(x) =2\int_{0}^{x}y\sin(2/y)\,dy-2G(x)$. Now divide by $x$ and let $x\to 0$. $\endgroup$ – Paramanand Singh Apr 25 at 10:42

I am posting an answer on the basis of hints given in comments, hope it helps if someone has the same question. Please tell if this is correct.

\begin{align*} \lim\limits_{x \to 0}\frac{1}{x} \int_{0}^{x} \sin^2\left( \frac{1}{u}\right) \mathrm du & = \lim\limits_{x \to 0}\frac{1}{x} \int_{0}^{x} \frac{\left(1-\cos\left( \frac{2}{u}\right)\right)}{2}\ \mathrm du\\ & = \lim\limits_{x \to 0}\left(\frac{1}{x}\cdot \frac{1}{2}\int_{0}^{x} du - \frac{1}{x}\cdot \frac{1}{2} \int_{0}^{x} \cos\left(\frac{2}{u} \right)\mathrm du \right)\\ & = \lim\limits_{x \to 0} \left(\frac{1}{2x}\cdot x \right) - \lim\limits_{x \to 0}\left( \frac{1}{2x} \int_{0}^{x} \cos \left( \frac{2}{u}\right)\mathrm du \right)\\ & = \frac{1}{2} - \frac{1}{2}\ \lim\limits_{x \to 0} \frac{1}{x} \int_{0}^{x} \cos \left( \frac{2}{u}\right) \mathrm du.\\ \end{align*}

{Claim} : $$ \lim\limits_{x \to 0} \frac{1}{x} \int_{0}^{x} \cos \left( \frac{2}{u} \right)\mathrm du = 0$$

{Proof} : Consider the function $F(x) = x^2\sin \left( \frac{2}{x} \right).$ Then,

$F(0) = 0$ as $\lim\limits_{x \to 0} x^2\sin \left(\frac{2}{x} \right) = 0\ (\because -1 \le \sin \left( \frac{2}{x} \right)\le 1 \Rightarrow -x^2 \le x^2 \sin \left( \frac{2}{x}\right) \le x^2.) $ Hence, we define function value to be $0$ at $x =0$ making $F(x)$ continuous at $x= 0$.

Also, we can calculate using limit definition that $F'(0) = 0 $ as follows :

\begin{align*} F'(0) & = \lim\limits_{x \to 0} \frac{F(x) - F(0)}{x-0}\\ & = \lim\limits_{x\to 0} \frac{x^2 \sin \left( \frac{2}{x}\right)}{x}\\ & = \lim\limits_{x \to 0} x \sin \left( \frac{2}{x}\right)\\ & = 0 \ \ \ \ \ \ \ \ \ \text{(using sandwitch theorem as done above).}\\ \end{align*}

We also know that $F'(x) = 2x \sin \left( \frac{2}{x} \right)- 2 \cos \left( \frac{2}{x}\right)$ whenever $x \ne 0$. To prove our claim, assume $$G(x) = \int_{0}^{x} \cos \left( \frac{2}{u} \right) \mathrm du$$

Now, by Fundamental theorem of calculus (FTOC), $$F(x) = \int_{0}^{x} F'(u) \mathrm du = 2 \int_{0}^{x} u\sin \left(\frac{2}{u}\right) \mathrm du - 2 G(x)$$

Next, dividing by $x$ and taking $\lim\limits_{x \to 0}$ we get, \begin{align*} \lim\limits_{x \to 0} \frac{F(x)}{x} & = \lim\limits_{x \to 0}\frac{2}{x} \int_{0}^{x} u \sin \left(\frac{2}{u}\right) \mathrm du - \lim\limits_{x \to 0} \frac{2}{x} G(x)\\ \Rightarrow F'(0) & = \lim\limits_{x \to 0}\frac{2}{x} \int_{0}^{x} u \sin \left(\frac{2}{u}\right) \mathrm du - \lim\limits_{x \to 0} \frac{2}{x} G(x)\\ \end{align*}

We see that $$\lim\limits_{x \to 0}\frac{2}{x} \int_{0}^{x} u \sin \left(\frac{2}{u}\right) \mathrm du = 0$$ To see this, we apply similar method as done above. Let $h(x) = x\sin \left(\frac{2}{x}\right).$ Since, $\lim\limits_{x \to 0}x\sin \left(\frac{2}{x} \right) = 0$, we define function value to be $0$ at $x =0$, making it continuous there. Next, let \begin{align*} H(x) &= \int_{0}^{x} h(u)\mathrm du\\ \Rightarrow \lim\limits_{x \to 0}\frac{H(x)}{x} &= \lim\limits_{x \to 0} \frac{1}{x}\int_{0}^{x} u\sin\left(\frac{2}{u} \right) \mathrm du\\ \Rightarrow H'(0) &= \frac{d}{dx} \int_{0}^{x} u\sin \left(\frac{2}{u} \right) \mathrm du\\ \Rightarrow H'(0) &= h(0) \ \ \ \ \ \text{(using FTOC)}\\ \Rightarrow H'(0) &= 0\\ \end{align*}

Therefore we have that,

\begin{align*} F'(0) &= 0- \lim\limits_{x \to 0} \frac{2}{x}G(x)\\ &\Rightarrow \lim\limits_{x \to 0} \frac{2}{x}G(x) = 0\\ &\Rightarrow \lim\limits_{x \to 0} \frac{1}{x} \int_{0}^{x} \cos \left( \frac{2}{u} \right)\mathrm du = 0.\\ \end{align*} Hence, the claim is proved. Next, using the claim we obtain finally, $$ \lim\limits_{x \to 0}\frac{1}{x} \int_{0}^{x} \sin^2\left( \frac{1}{u}\right) \mathrm du = \frac{1}{2}\ .$$

  • $\begingroup$ This is exactly what I mentioned in comments. +1 for getting it right. A few minor things to be fixed here. We define $F(0)=0$ so that $F$ becomes continuous at $0$. Next the limit of $(1/x)\int_{0}^{x}u\sin(2/u)\,du$ is $0$ using similar mechanism. Let $h(x) =x\sin(2/x),h(0)=0, H(x) =\int_{0}^{x}h(u)\,du$ then $h$ is continuous at $0$ and by FTC we have $H'(0)=h(0)=0$. The use of L'Hospital's Rule here is pretty roundabout. $\endgroup$ – Paramanand Singh Apr 27 at 14:47
  • $\begingroup$ You should have a look at math.stackexchange.com/a/2603106/72031 $\endgroup$ – Paramanand Singh Apr 27 at 14:56
  • $\begingroup$ And also see a harder problem which requires different technique math.stackexchange.com/q/2063078/72031 $\endgroup$ – Paramanand Singh Apr 27 at 14:58
  • $\begingroup$ Thanks for the links. $\endgroup$ – Rick Apr 27 at 15:04
  • $\begingroup$ @ParamanandSingh Also, I have already written that $F(0) = 0$ in the beginning, is that correctly defined? $\endgroup$ – Rick Apr 27 at 15:16

I could be wrong but I think that the limit is undefined:

Let $g'(y) = \sin^2(\frac{1}{y})$ and $f(x) = \frac{1}{x} \int_0^x \sin^2(\frac{1}{y}) dy$.

Then $f(x) =\frac{1}{x} \int_0^x g'(y) dy$

$=\frac{1}{x} [g(y)]^x_0$

$= \frac{g(x) - g(0)}{x}$

Using L'Hopital's rule:

$\lim_{x\to \infty} f(x) = \frac{g'(x)}{1} = g'(0)$, which is undefined.

  • $\begingroup$ I think there is some fault as you multiply by $x$ freely, although it can be equal to $0$? The whole problem is when $x \rightarrow 0$. The solution given by @Paramanand Singh works without assuming function value at $0$. $\endgroup$ – Rick Apr 29 at 7:16
  • $\begingroup$ I have edited my answer in a way that does not involve multiplying by $x$ and i haven't assumed the value of any functions at 0. $\endgroup$ – 1123581321 Apr 29 at 7:18
  • $\begingroup$ I think now I get it, you integrate $g'(y)$ from $0$ to $x$, but it is not continuous at $x= 0$? So does the integration work? $\endgroup$ – Rick Apr 29 at 7:20
  • $\begingroup$ Wouldn't that imply that the limit is undefined then? $\endgroup$ – 1123581321 Apr 29 at 7:21
  • 2
    $\begingroup$ Your approach has a flaw. The integrand is discontinuous at $0$ and hence the fundamental theorem of calculus does not apply. $\endgroup$ – Paramanand Singh Apr 29 at 7:32

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