How can I calculate the following limit?

$$\begin{equation*} \lim_{x \rightarrow \infty} (x + \sin x)\sin \frac{1}{x} \end{equation*},$$

First, I multiplied and then distributed the limit then the limit of the first term was 1 but the second term was $\sin (x) \sin (1/x)$, I used the rule $\sin x \sin y = 1/2\{\cos(x-y) - \cos(x+y)\}$, but I got stucked, any help will be appreciated.


  • 3
    $\begingroup$ Note that $\sin(x)$ is bounded and $\sin(1/x)\to 0$ as $x\to\infty$. $\endgroup$ – Robert Z Nov 9 '17 at 15:46
  • $\begingroup$ yeah brilliant thank you! @RobertZ $\endgroup$ – Emptymind Nov 9 '17 at 15:53
  • $\begingroup$ @RobertZ : so it becomes the famous $\displaystyle\lim_{x\to\infty}x\sin(x^{-1})$ $\endgroup$ – mathreadler Nov 9 '17 at 15:54

To find this limit, you're going to need to employ the following two well-known facts from the theory of limits:

$$ \lim_{\theta \rightarrow 0}\frac{\sin\theta}{\theta}=1 $$

and $$ \lim_{\theta \rightarrow \infty}\frac{\sin\theta}{\theta}=0 $$

The proofs of those two statements are here and here respectively.

Notice that as $x$ approaches infinity, $\frac{1}{x}$ approaches $0$. So, $x\rightarrow \infty\implies\frac{1}{x}\rightarrow 0$. Also, don't forget that $x$ is the same thing as $\frac{1}{\frac{1}{x}}$ as long as $x$ does not equal zero (it doesn't in our case here):

$$ \lim_{x \rightarrow \infty}\left[(x + \sin x)\cdot\sin\frac{1}{x}\right]= \lim_{x \rightarrow \infty}\left(x\cdot\sin\frac{1}{x}\right) +\lim_{x \rightarrow \infty}\left(\sin x\cdot\sin\frac{1}{x}\right)=\\ \lim_{\frac{1}{x} \rightarrow 0}\frac{\sin\frac{1}{x}}{\frac{1}{x}}+ \lim_{x \rightarrow \infty}\left(\frac{x}{x}\cdot\sin x\cdot\sin\frac{1}{x}\right)=\\ 1+\lim_{x \rightarrow \infty}\frac{\sin x}{x}\cdot \lim_{\frac{1}{x}\rightarrow 0}\frac{\sin \frac{1}{x}}{\frac{1}{x}}=\\ 1 + 0\cdot 1=1 $$


$$\lim_{x \rightarrow \infty} (x + \sin x)\sin \frac{1}{x}=\lim_{x\rightarrow\infty}\frac{\sin\frac{1}{x}}{\frac{1}{x}}+0=1$$

  • $\begingroup$ but the second term is $0.\infty$ which as an indeterminate form? $\endgroup$ – Emptymind Nov 9 '17 at 15:52
  • 1
    $\begingroup$ @Intuition $|\sin{x}|\leq1$ and $\sin\frac{1}{x}\rightarrow0.$ $\endgroup$ – Michael Rozenberg Nov 9 '17 at 16:07

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