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Given the problem:

$\lim_{x\to \infty} x\sin({1\over x})$

I am confused as to how to obtain the given answer of $1$. My steps are as follows: First I moved the $x$ to the denominator to become

$\lim_{x\to \infty}{\sin{1\over x}\over {1\over x}}$

Next, I set ${1\over x}$ equal to $u$ to get $\lim_{x\to \infty}$ of ${{\sin (u) \over u}}$

However, I am unsure how to proceed. Plugging in infinity seems to be the logical step, but I am still unsure if that would yield anything helpful. Any advice would be appreciated.

Note: I would like to just forwarn that I am a high school junior just starting AP Calc BC so I am sorry if my question seems a bit trivial

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mlz7
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    You'll have to change the limit to something involving $u$. As $x\to \infty$, to what does $u$ tend? –  Sep 26 '16 at 22:34
  • @Bye_World I am unsure how to answer this. I know that the limit as ${u\to \0} of that would be 1, but I am unsure how to manipulate the step I left off at in order to resemble this. – mlz7 Sep 26 '16 at 22:46
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    You define $u$ by $u=\frac 1x$. Then take the limit. As $x\to \infty$ we have $u\to 0^+$. Thus, doing the change of coordinates: $$\lim_{x\to \infty} \frac{\sin\left(\frac 1x\right)}{\frac 1x} = \lim_{u\to 0^+} \frac{\sin(u)}{u}$$ which as you've rightly said is equal to $1$. –  Sep 26 '16 at 22:51
  • @Bye_World thank you so much. I was initially confused as to how you were able to go from finding the limit as x approaches infinity to finding the limit as u approaches zero but I believe I understand it now. I would mark your comments as the answer if it were possible. – mlz7 Sep 26 '16 at 23:09

2 Answers2

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Set

$$x\to \frac{1}{y}$$

Hence

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

Enrico M.
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Rather than change variables, one can simply recall from elementary geometry (SEE THIS ANSWER)for $|\theta|\le \pi/2$, we have

$$\theta\cos(\theta)\le\sin(\theta)\le \theta \tag 1$$

Using $(1)$ with $\theta =1/x$ for $x>0$ reveals

$$\cos(1/x)\le \frac{\sin(1/x)}{1/x}\le 1 \tag 2$$

Applying the squeeze theorem and exploiting the continuity of the cosine function yields the coveted limit

$$\lim_{x\to \infty}\frac{\sin(1/x)}{1/x}=1$$

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Mark Viola
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