This in an exercise in my Analysis book in a section on L'Hopital's rules. $$\lim\limits_{x \to \infty} \left[\sin\left(\frac{1}{x}\right)\right]^x$$ Now it's an indeterminate of the form $0^\infty$ however I don't know how to solve this. I have tried the following:

$$y=\lim\limits_{x \to \infty} \left[\sin\left(\frac{1}{x}\right)\right]^x$$ $$\ln y=\lim\limits_{x \to \infty}\ln \left[\sin\left(\frac{1}{x}\right)\right]^x$$ $$\ln{y}=\lim\limits_{x \to \infty} x\ln{\sin\frac{1}{x}}$$ Now this is an indeterminate limit of form $\infty\cdot\infty$ which approaches $\infty$. However I may not write now that therefore $y=e^\infty=\infty$.

How do I write this out correctly?

  • 4
    $\begingroup$ Actually $\ln \sin 1/x$ approaches to $- \infty$, so that the original limit is $0$. Anyway, $0^{\infty}$ is not an indeterminate form: it is always $0$. $\endgroup$
    – Crostul
    Commented Dec 18, 2016 at 0:04
  • 1
    $\begingroup$ $\sin\left(\dfrac{1}{x}\right)$ is equivalent to $\dfrac{1}{x}$ as $x$ approaches $\infty$. $\endgroup$
    – drzbir
    Commented Dec 18, 2016 at 0:10
  • 1
    $\begingroup$ you are really a Question Maker .......+1 $\endgroup$ Commented Dec 18, 2016 at 5:54
  • $\begingroup$ Related: $\displaystyle\lim_{x \to \infty} \, \cos \left(\dfrac{1}{x}\right)^{x}$. $\endgroup$
    – Workaholic
    Commented Dec 18, 2016 at 9:48

4 Answers 4



In THIS ANSWER, I developed the pair inequalities that is introduced in elementary geometry

$$\bbox[5px,border:2px solid #C0A000]{\theta\cos(\theta) \le \sin(\theta) \le \theta} \tag 1$$

for $0\le \theta \le \pi/2$.

Using $(1)$ with $\theta=1/x$, we have

$$\frac{\cos(1/x)}{x}\le \sin(1/x)\le \frac1x$$

for $\frac2\pi \le x$. Hence, we can write

$$\left(\cos(1/x)\right)^{x} \frac{1}{x^x}\le \left(\sin(1/x)\right)^x\le \frac1{x^x}$$

whence application of the squeeze theorem yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty} \left(\sin(1/x)\right)^x=0}$$

  • $\begingroup$ pretty nice squeezing there! $\endgroup$
    – user459879
    Commented Feb 20, 2019 at 19:39
  • 1
    $\begingroup$ Thank you Wesley. Much appreciated. $\endgroup$
    – Mark Viola
    Commented Feb 20, 2019 at 23:52
  • $\begingroup$ Woudl the downvoter care to comment on the reason that this is not useful? $\endgroup$
    – Mark Viola
    Commented Sep 2, 2022 at 23:40

For $x>6/\pi,$ $0<\sin(1/x) <1/2.$ For such $x$ we have $0 < \sin(1/x)^x < (1/2)^x \to 0.$ Therefore the limit is $0$ by the squeeze theorem.


Substitute $x=1/t$ and compute the limit of the logarithm: $$ \lim_{x\to\infty}\log\left(\left((\sin\frac{1}{x}\right)^x\right) =\lim_{x\to\infty}x\log\left(\sin\frac{1}{x}\right) =\lim_{t\to0^+}\frac{\log\sin t}{t}=-\infty $$ Thus your limit is $\lim_{u\to-\infty}e^u=0$.

  • $\begingroup$ the OP wrote $\sin\left( \frac{1}{x}\right)^x$? You mean, this one is the same with the one you wrote above? $\endgroup$ Commented Dec 18, 2016 at 0:19
  • $\begingroup$ @juniven Look the last step in the question. $\endgroup$
    – egreg
    Commented Dec 18, 2016 at 0:31
  • $\begingroup$ ah ok. Maybe I have to edit the problem $\endgroup$ Commented Dec 18, 2016 at 0:32
  • $\begingroup$ Any way of proving this without using equivalence characteristics? $\endgroup$ Commented Dec 18, 2016 at 1:51

In fact you already know that $|\sin(u)|\le1$ for all $u$, and when $u\to0$ this sinus also gets smaller.

So no matter what's really inside the sinus, it can be a much more complicated expression than just $u=1/x$, till it can become sufficiently small then you have :

$|\sin(u)|\le C<1$ and it is well known that $C^x\to0$ when $x\to+\infty$ and so is the whole expression.

For instance if you ask me the same question for $\sin(\frac{53+x^2-\ln(x)}{\sqrt{\pi^x}})^x$ then my answer will be exactly the same.

And it doesn't even need to tend towards $0$, we just need the quantity inside the sinus to be small enough [ i.e. $|u|\le\alpha$ with $\alpha<\frac{\pi}{2}$ so that $C<1$]. For instance $sin(\frac{3x^2+5x+31}{2x^2-7})^x$ will do.

Of course you can always do finer majorations, but being lazy in mathematics is not always bad ;


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .