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Find $$\lim_{x \to \infty} x\sin\left(\frac{11}{x}\right)$$

We know $-1\le \sin \frac{11}{x} \le 1 $

Therefore, $x\rightarrow \infty $ And so limit of this function does not exist.

Am I on the right track? Any help is much appreciated.

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It is true that

$$-1\le \sin \frac{11}{x} \le 1$$

but since $x\to \infty$ we have that $$\sin \frac{11}{x}\to0$$

thus the limit is in the indeterminate form $0\cdot \infty$.

To solve we can set for example $y=\frac1x\to 0$ then

$$\lim_{x \to \infty} x\sin\left(\frac{11}{x}\right)=\lim_{y \to 0} \frac{\sin\left(11y\right)}{y}=11\cdot\lim_{y \to 0} \frac{\sin\left(11y\right)}{11y}=11$$

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  • $\begingroup$ Since the denominator is 11y and y goes to zero, 11y should be zero. And we cannot divide by 0. I am sorry I think I am not getting the intermediate step. $\endgroup$ – nova_star Feb 21 '18 at 22:17
  • $\begingroup$ @nova_star Note that for definition of limit $y\to 0$ doesn't mean tha $y=0$ otherwise it is assumed that $y\neq0$. Refer also to this related OP math.stackexchange.com/questions/2628911/… $\endgroup$ – gimusi Feb 21 '18 at 22:22
  • $\begingroup$ Now this makes sense. You are using the fact that limit of sin(11y)/11y is 1 when y goes to 0. $\endgroup$ – nova_star Feb 22 '18 at 11:44
  • $\begingroup$ Yes of course! Exactly $\endgroup$ – gimusi Feb 22 '18 at 11:49
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$$\sin(11/x)\underset{(+\infty)}{\sim}11/x$$ What can you deduce ?

Note : What you have stated is good however with the product with $x$ you cannot conclude that it converges or diverges with what you wrote

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write $$11\frac{\sin(\frac{11}{x})}{\frac{11}{x}}$$ and the Limit is $$11$$

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  • $\begingroup$ As x approaches infinity 11/x should approach zero. So we will have zero as the denominator. How is the limit 11 then? Sorry, having a hard time understanding. And appreciate the help. $\endgroup$ – nova_star Feb 21 '18 at 21:46
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    $\begingroup$ you have the Limit $$\frac{\sin(u)}{u}$$ as $u$ tends to Zero! $\endgroup$ – Dr. Sonnhard Graubner Feb 21 '18 at 21:48
  • $\begingroup$ since $$\frac{11}{x}$$ tends to Zero as $x$ tends to infinity $\endgroup$ – Dr. Sonnhard Graubner Feb 21 '18 at 21:49
  • $\begingroup$ see the other post, he/she has written the same! $\endgroup$ – Dr. Sonnhard Graubner Feb 21 '18 at 21:50
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Hint

$$\lim\limits_{x \to \infty} x\sin\left(\frac{11}{x}\right)=11\lim\limits_{x \to 0^+} \frac{\sin(11x)}{11x}$$

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No the reasoning doesn't follow. If limit exists, then using your reasoning all we can say is it is between $-\infty$ and $\infty$. Make the change of variables $u=1/x$, and note that the limit is equivalent to $$ \lim_{u\to 0^+}\frac{\sin 11u}{u}=11\lim_{u\to 0^+}\frac{\sin 11u}{11 u} $$ and now use the well-known limit $$ \lim_{x\to 0}\frac{\sin x}{x}=1 $$

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Your conclusion is not directly correct, since you are neglecting the $x$ in the denominator inside the $\sin \frac{11}{x}$.

Write

$$ \lim_{x \rightarrow \infty} x \sin \frac{11}{x} = \lim_{x \rightarrow \infty} \frac {\sin \frac{11}{x}}{\frac{1}{x}}. $$

and L'Hôpital's rule is applicable.

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