I'm trying to understand how

$$\lim_{n\rightarrow\infty}\left(1-\sin{\frac1 n}\right)^n$$

is equal to $e^{-1}=\frac 1 e$. I know that

$$e = \lim_{n\rightarrow\infty}\left(1+\frac 1 n\right)^n$$

and that $$\sin\frac 1 n \sim \frac 1 n,$$

so theoretically I should be able to do something like this:

$$\lim_{n\rightarrow\infty}\left(1-\sin{\frac1 n}\right)^n =\lim_{n\rightarrow\infty}e^{\ln(1-\sin{\frac1 n})^n}= \cdots$$

But I just don't know how to go from there without the Hôpital rule. Any hints?

  • 1
    $\begingroup$ You're taking a limit on $x$ of a function that doesn't have $x$ in it? $\endgroup$ – Gerry Myerson Nov 7 '17 at 12:02
  • $\begingroup$ Wow sorry, editing $\endgroup$ – Cesare Nov 7 '17 at 12:03
  • $\begingroup$ Do you also know that $(1+\alpha/n)^n\to e^{\alpha}$? Use $\alpha=-1$ for your case. $\endgroup$ – M. Winter Nov 7 '17 at 12:05
  • $\begingroup$ Thanks @M.Winter! Feel free to post that as an answer and I'll be happy to accept it. $\endgroup$ – Cesare Nov 7 '17 at 12:10
  • 2
    $\begingroup$ For every $(x_n)$ such that $$nx_n\to x$$ one has $$(1+x_n)^n\to e^x$$ This has ben explained tons of times on the site. $\endgroup$ – Did Nov 7 '17 at 12:11

Note the more general formula

$$\lim_{n\to\infty}\left(1+\frac {\color{red}\alpha} n\right)^n=e^{\color{red}\alpha}.$$

In your case you can use $\alpha=-1$ to find $(1-1/n)^n\to e^{-1}$. From this we have

$$ \lim_{n\to\infty}\left(1-\sin\frac 1n\right)^n =\lim_{n\to\infty}\left(1-\frac 1n\right)^n =e^{-1}. $$

Note that the first equals sign might need some more justification (which you might find in other answers or @Did's comment as well).

  • $\begingroup$ How do you justify $\lim_{n\to\infty}\left(1-\sin\frac 1n\right)^n =\lim_{n\to\infty}\left(1-\frac 1n\right)^n$ ? Is $$\lim_n \left(\frac{1-\sin{\frac1 n}}{1-\frac1n}\right)^n = 1$$ trivial ? $\endgroup$ – Gabriel Romon Nov 7 '17 at 12:14
  • $\begingroup$ @GabrielRomon You are right about this gap. I just wanted to clarify some part of Andrea's answer. If this is a specific cause of trouble for OP then I will think about an explanation. Or OP can look at your answer ;) $\endgroup$ – M. Winter Nov 7 '17 at 12:18
  • $\begingroup$ @GabrielRomon: you can establish this using a lemma (first popularized on this website by user Thomas Andrews) : if $n(a_{n} - 1)\to 0$ then $a_{n} ^{n} \to 1$. Now check this with $a_{n} =(1-\sin(1/n))/(1-(1/n))$. $\endgroup$ – Paramanand Singh Nov 7 '17 at 13:48

$$ \lim_{n\rightarrow\infty}(1-\sin{\frac1 n})^n = \lim_{n\rightarrow\infty}(1-\frac1n)^n = e^{-1} $$

The last step can be seen by letting $m = -n$, giving

$$ (\lim_{m\rightarrow-\infty}(1+\frac1m)^m)^{-1} = e^{-1} $$

  • $\begingroup$ Thanks. Where do you get the -1 from? $\endgroup$ – Cesare Nov 7 '17 at 12:05
  • 1
    $\begingroup$ I don't think it's that simple... You need to justify $$\lim_n \left(\frac{1-\sin{\frac1 n}}{1-\frac1n}\right)^n = 1$$ $\endgroup$ – Gabriel Romon Nov 7 '17 at 12:09
  • $\begingroup$ I put in in the text. $\endgroup$ – Andreas Nov 7 '17 at 12:09
  • $\begingroup$ @GabrielRomon For large n, the next term in the sin will be $\propto 1/n^3$ which will not cause any change. $\endgroup$ – Andreas Nov 7 '17 at 12:12

Maybe OP doesn't know small oh notation yet, but still: $$\begin{aligned}[t]\left( 1-\sin{\frac1 n}\right)^n &= \exp\left(n\ln\left(1-\sin \frac 1n \right)\right)\\ &= \exp\left(n\ln\left(1-\frac 1n + o\left(\frac 1n \right) \right)\right)\\ &=\exp(-1+o(1))\\ &= e^{-1}+o(1) \end{aligned}$$


Taking log you obtain that $n \log (1-\sin \frac{1}{n})=\frac{\log (1-\sin \frac{1}{n})}{\frac{1}{n}}$. Using L'hopital you get $\frac{\log (1-\sin \frac{1}{n})}{\frac{1}{n}}=\frac{-\cos\frac{1}{n}}{1-\sin\frac{1}{n}}\rightarrow -1$ as $n\to\infty$. Therefore $(1-\sin\frac{1}{n})^n$ converges to $e^{-1}$.

Here is an alternative proof without using L'hopital, but instead using Taylor: again taking log you obtain $\frac{\log (1-\sin \frac{1}{n})}{\frac{1}{n}}$. Expand $\log (1-\sin \frac{1}{n})=\log 1-\frac{1}{1+o(1)}\sin \frac{1}{n}$ by using Taylor, we obtain that $\frac{\log (1-\sin \frac{1}{n})}{\frac{1}{n}}=\frac{-\frac{1}{1+o(1)}\sin \frac{1}{n}}{\frac{1}{n}}$, therefore we only need to show that $\frac{\sin x }{x}$ converges to one as $x\to 0$. But again from Taylor $\sin x=\sin 0+\cos (o(x)) x$, therefore $\frac{\sin x }{x}=\cos (o(x))$ converges to one as $x\to 0$.

  • $\begingroup$ Cheers, unfortunately I'm not allowed to use L'hopital rule. $\endgroup$ – Cesare Nov 7 '17 at 12:12
  • $\begingroup$ ok o.O, then let me think about another method :) $\endgroup$ – Student Nov 7 '17 at 12:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.