Find $\lim\limits_{x \to \infty} x\sin\frac{11}{x}$

Question

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.

Answer 1

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

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

Answer 2

$$\sin(11/x)\underset{(+\infty)}{\sim}11/x$$ 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

Answer 3

Hint

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

Answer 4

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 $$ \lim_{x\to 0}\frac{\sin x}{x}=1 $$

Answer 5

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

Answer 6

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.