here's the question, how can I solve this:

$$\lim_{x \rightarrow \infty} x\sin (1/x) $$

Now, from textbooks I know it is possible to use the following substitution $x=1/t$, then, the ecuation is reformed in the following way

$$\frac{\sin t}{t}$$

then, and this is what I really can´t understand, textbook suggest find the limit as $t\to0^+$ (what gives you 1 as result)

Ok, I can't figure out WHY finding that limit as $t$ approaches $0$ from the right gives me the answer of the limit in infinity of the original formula. I think I can't understand what implies the substitution.

Better than an answer, I need an explanation.

(Sorry If I wrote something incorrectly, the english is not my original language) Really thanks!!

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    $\begingroup$ The limit you wrote does not exist.The limit of $x\sin(1/x)$ as $x\to\infty$ is $1$, for the reason you describe. $\endgroup$ – Andrés E. Caicedo Jul 20 '14 at 4:30
  • $\begingroup$ Did you mean $\displaystyle\lim_{x \to \infty}x \sin \dfrac{1}{x} = 1$? $\endgroup$ – JimmyK4542 Jul 20 '14 at 4:30
  • $\begingroup$ Yes, Im sorry, I mean lim x sin (1/x) as x tends to infinity just as you said. Thanks $\endgroup$ – Matías Leonardo Nieto Jul 20 '14 at 4:35
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    $\begingroup$ The books do not suggest to compute the limit of $\sin t/t$ as $x\to 0$. Remember that $x\to\infty$ and $t=1/x$, so $t\to 0$. You need to study $\displaystyle\lim_{t\to 0}\frac{\sin t}t$. And this is a standard limit that your book probably covers in detail. $\endgroup$ – Andrés E. Caicedo Jul 20 '14 at 4:40
  • $\begingroup$ Understood Andres, yes I know why (sin t)/t is 1 as x tends to 0. I couldn't understand how the substitution worked with the limit. Thanks!! $\endgroup$ – Matías Leonardo Nieto Jul 20 '14 at 4:44

$$\lim_{x\to\infty}x\sin(1/x)$$ is like the limit of this sequence: $$1\sin(1/1), 10\sin(1/10), 100\sin(1/100),\ldots$$ where we have inserted $x$ marching along from $1$ to $10$ to $100$ on its merry way to $\infty$. This is almost literally the same as $$\frac{1}{1}\sin(1), \frac{1}{0.1}\sin(0.1), \frac{1}{0.01}\sin(0.01),\ldots$$ which is an interpretation of $$\lim_{t\to0^+}\frac{1}{t}\sin(t)$$ with $t$ marching its way from $1$ down to $0.1$ down to $0.01$ on its merry way to $0$. So whatever the value of these limits are, $\lim\limits_{x\to\infty}x\sin(1/x)=\lim\limits_{t\to0^+}\frac{1}{t}\sin(t)$.

  • $\begingroup$ While this is a good intuitive explanation, an important thing to note is that $x \to +\infty$ is equivalent to $\frac{1}{x} \to 0^+$. That is the key point. $\endgroup$ – user21820 Jul 20 '14 at 6:31
  • $\begingroup$ @user21820 I felt that if OP did not understand that yet, then reiterating it (from the comments, and no doubt the textbooks) was not going to help. This answer is more to help OP understand something than to give a mathematically rigorous response. $\endgroup$ – alex.jordan Jul 20 '14 at 6:38
  • $\begingroup$ Yes I understood that, which is why I upvoted your answer. But after the intuition should come the verification. =) $\endgroup$ – user21820 Jul 20 '14 at 6:40
  • $\begingroup$ @user21820 Someone down voted; can't understand why, but such is down voting here sometimes :) $\endgroup$ – alex.jordan Jul 20 '14 at 6:40
  • $\begingroup$ So you thought it was me? Haha I downvote only if it is not an answer, useless or there is an actual conceptual error. =) $\endgroup$ – user21820 Jul 20 '14 at 6:41

You can simply observe that $$ \lim_{x\rightarrow+\infty}x\sin(1/x)= \lim_{x\rightarrow+\infty}x\frac{1}{x}\frac{\sin(1/x)}{\frac{1}{x}}= \lim_{x\rightarrow+\infty}\frac{\sin(1/x)}{1/x} $$

Now it should be clear that $1/x$ tends to $0$ as $x$ approaches to $+\infty$. Hence it will be equivalent to write $1/x=t$ and let $t$ tends to $0$. Thus you have $$ \lim_{x\rightarrow+\infty}\frac{\sin(1/x)}{1/x}= \lim_{t\rightarrow0}\frac{\sin t}{t}=1\;, $$ as wanted.

  • $\begingroup$ It's equivalent but not for the reason you give. Specifically $x$ may not tend to $\infty$ as $\frac{1}{x}$ tends to $0$. So your reasoning is enough to conclude that if the limit exists as $\frac{1}{x} \to 0$, then the limit exists as $x \to \infty$, but not conversely. $\endgroup$ – user21820 Jul 20 '14 at 6:25
  • $\begingroup$ Thanks. Can you explain why $\frac1{x}$ tends to $\infty$ doesn't imply necessarely that $x$ tends to $0$? Thanks again $\endgroup$ – Joe Jul 20 '14 at 8:15
  • $\begingroup$ Please read a bit more carefully. I said that the limit may exist as $x \to \infty$ but may not exist as $\frac{1}{x} \to 0$. Consider that $x$ can be negative in the second case. $\endgroup$ – user21820 Jul 20 '14 at 9:17
  • $\begingroup$ In fact I shouldn't have said that it's "equivalent"; it's not, because of what I've said. I meant to say that the limits are equal in this case. $\endgroup$ – user21820 Jul 20 '14 at 9:18

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