# Solve $\lim_{n\rightarrow\infty}e^{\ln(1-\sin{\frac1 n})^n}$

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?

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

• 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 ? – Gabriel Romon Nov 7 '17 at 12:14
• @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 ;) – M. Winter Nov 7 '17 at 12:18
• @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 $a_{n} =(1-\sin(1/n))/(1-(1/n))$. – 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

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

• Thanks. Where do you get the -1 from? – Cesare Nov 7 '17 at 12:05
• I don't think it's that simple... You need to justify $$\lim_n \left(\frac{1-\sin{\frac1 n}}{1-\frac1n}\right)^n = 1$$ – Gabriel Romon Nov 7 '17 at 12:09
• I put in in the text. – Andreas Nov 7 '17 at 12:09
• @GabrielRomon For large n, the next term in the sin will be $\propto 1/n^3$ which will not cause any change. – 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 $x\to 0$.

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