Limit involving $x$ raised to $\sin (1/x)$

Question

the limit as $x \to \infty$ of $x ^{\sin(1/x)}$

I see this is an infinity to zero situation, and I so I know this will be an $e^{\text{something}}$ sort of answer and that my next step would look something like: limit as $x$ goes to infinity of $\ln(x^{\sin(1/x)})$... what is my next step and how do I proceed?

Thanks much!

Hi Kavi, there was a mistake in my original post. Apologies. — jackbenimbo, Aug 03 '18 at 06:44
@jackbenimbo What is the argument for sine? You should edit your question, it is not clear what youare asking for. — user, Aug 03 '18 at 06:48
Hi @gimusi, I found the correct problem. My greatest apologies. — jackbenimbo, Aug 03 '18 at 06:50
@jackbenimbo Also $\sin*(1/x)$ is meaningless, maybe it is $\sin(1/x)$? — user, Aug 03 '18 at 06:52
@jackbenimbo Your idea to use logarithm is a correct way, I've indicated a way to proceed by that. — user, Aug 03 '18 at 07:00
@jackbenimbo Refer to standard limit whenever you can and do not use l'Hopital too much for any kind of limit! Bye — user, Aug 03 '18 at 07:03
@jackbenimbo Also follow the suggestion given here for the next questions https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question and learn https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference — user, Aug 03 '18 at 07:05
Will do. One quick question which may not illicit a quick response..for the limits I've worked with thus far in calc 1/2, LH has been a life saver. Why do you recommend against it? — jackbenimbo, Aug 03 '18 at 07:06
@jackbenimbo To learn limits it is important to recognize the standard limit as for example as $x\to 0$ $\sin x/x=1$, $(1-\cos x)/x^2=1/2$, $x\log x=0$ which can be proved independently (of course you should study these proof). L'Hopital is nice for some situation but by the blind use of it you can't learn to manipulate properly the limit by that. A better tool is Taylor's expansion, maybe you'll learn it later. Bye — user, Aug 03 '18 at 07:10

user · Accepted Answer · 2018-08-03T07:01:17.003

2

We have that

$$x ^{\sin(1/x)}=e^{\sin \left(\frac1x\right) \log x}=e^{\frac{\sin \left(\frac1x\right)}{\frac1x} \cdot \frac{\log x}{x}}\to e^{1 \cdot 0}=e^0=1$$

indeed since $t=\frac1x \to 0$

$$\frac{\sin \left(\frac1x\right)}{\frac1x}=\frac{\sin t}{t}\to 1$$

and it easy to show (for example by l'Hopital)

$$\frac{\log x}x \to 0$$

edited Aug 03 '18 at 07:01

answered Aug 03 '18 at 06:44

user

154,566

2

Another way to show that $(\log x)/x\to 0:$ For $1<n<x$ we have $\log x$ $=\int_1^x(1/y)dy=$ $\int_1^{x/n}(1/y)dy+\int_{x/n}^x(1/y)dy<$ $\int_1^{x/n}1dy+\int_{x/n}^x (n/x)dy=$ $=(x/n-1)+(x-x/n)(n/x)=(x/n)+n-2. $ So for all $x>n^2>1$ we have $0<(\log x)/x <$ $<((x/n)+n-2)/x=$ $1/n+(n-2)/x<$ $2/n.$ – DanielWainfleet Aug 04 '18 at 05:23
@DanielWainfleet Yes of course there are best way I’ve suggested l’Hopital in that case since it is straightforward and moreover this is a standardlimit which should well known and already proved. Another (basic) way is by $x=e^y$ and then $y/e^y\to 0$ from $n/e^n\to 0$ which can proved by induction. Thanks – user Aug 04 '18 at 06:20

score 2 · Answer 2 · answered Aug 03 '18 at 07:40

Consider $$y=x^{\sin \left(\frac{1}{x}\right)}\implies \log(y)={\sin \left(\frac{1}{x}\right)}\log(x)$$ Now, use Taylor expansion for large $x$ (or use equivalents) $${\sin \left(\frac{1}{x}\right)}=\frac{1}{x}+O\left(\frac{1}{x^3}\right)$$ then $$\log(y) \sim \frac{\log(x)}x\to 0 \qquad \text{making}\qquad y \to 1$$

score 1 · Answer 3 · answered Aug 03 '18 at 07:27

1

Alternatively, note: $\sin \frac 1x<\frac 1x, x>0$. Then: $$\left(\sin \frac 1x\right)^{\sin \frac 1x}<\left(\frac 1x\right)^{\sin \frac 1x}<x^{\sin \frac 1x}<x^{\frac 1x}, x>1$$ Taking limit at $x\to\infty$: $$1\le \lim_{x\to\infty} x^{\sin \frac 1x}\le 1.$$

Also: See this link.

answered Aug 03 '18 at 07:27

farruhota

31,482

This is an awesome shortcut! Thanks so much – jackbenimbo Aug 03 '18 at 07:31
1

You are welcome. Good luck. – farruhota Aug 03 '18 at 07:32

Limit involving $x$ raised to $\sin (1/x)$

3 Answers3