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I've been trying to solve this limit: $$\lim_{n\to \infty} \left(\cos\left(\frac{1}{n}\right)\right)^{n^2}$$ I solved it using l'hopital but I have to try in another way. I tried to make the expression: $$ \lim_{n \to \infty}\exp\left(n^2\ln\left(\cos\left(\frac{1}{n}\right)\right)\right)$$

but it didn't work out for me...

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    $\begingroup$ $\lim_{x\to\infty} \cos\left(\frac{1}{n}\right)^{n^2} = \cos\left(\frac{1}{n}\right)^{n^2}$, since it doesn't depend on $x$. Did you mean to write $\lim_{n\to\infty}$? $\endgroup$
    – Xander Henderson
    Sep 16, 2017 at 14:08
  • $\begingroup$ Your'e absolutely right. edited.. thanks! $\endgroup$
    – LosLas
    Sep 16, 2017 at 14:09
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    $\begingroup$ Maybe a better title is in order? Something... that might reflect the actual content of the question? $\endgroup$
    – Asaf Karagila
    Sep 16, 2017 at 14:20
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    $\begingroup$ See what @Asaf meant? $\endgroup$
    – Did
    Sep 16, 2017 at 14:27

6 Answers 6

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$$ \lim_{n\to\infty}n^2\ln\left(\cos{1\over n}\right)= \lim_{n\to\infty}{\ln\left(1-\left(1-\cos{1\over n}\right)\right) \over1-\cos{1\over n}}\cdot{1-\cos{1\over n}\over(1/n)^2}= -1\cdot{1\over2}=-{1\over2}. $$

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    $\begingroup$ You have a typo. The last expression should be $(-1)\cdot\frac{1}{2}$ as the log limit is $-1$. $\endgroup$
    – Paramanand Singh
    Sep 16, 2017 at 14:38
  • $\begingroup$ Right, thanks! Corrected now. $\endgroup$ Sep 16, 2017 at 15:23
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How about this \[ \cos(1/n)=1-\frac{1}{2n^2}+O\left(\frac{1}{n^4}\right) \] Then use the relation \[\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^n=e^x\] Combining the two one gets \[ \lim_{n\rightarrow\infty}[\cos(1/n)]^{n^2}=\lim_{n\rightarrow\infty}\left(1-\frac{1}{2n^2}\right)^{n^2}=e^{-1/2}=\frac{1}{\sqrt{e}} \]

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  • $\begingroup$ I only saw your answer after I posted mine. $\endgroup$
    – minmax
    Sep 16, 2017 at 14:25
  • $\begingroup$ No worry. That has happened to me a number of times. $\endgroup$
    – Mark Viola
    Sep 16, 2017 at 14:31
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Using $\cos(1/n)=1-\frac{1}{2n^2}+O\left(\frac1{n^4}\right)$, we have

$$\begin{align} \cos^{n^2}(1/n)&=\left(1-\frac{1}{2n^2}+O\left(\frac1{n^4}\right)\right)^{n^2}\\\\ &=\left(1-\frac{1}{2n^2}\right)^{n^2}\,\left(1+\frac{O\left(\frac{1}{n^4}\right)}{1-\frac1{2n^2}}\right)^{n^2} \end{align}$$

Next, we note that $\lim_{n\to \infty}\left(1-\frac{1}{2n^2}\right)^{n^2}=e^{-1/2}$ from the definition of the exponential function.

Finally, inasmuch as

$$\lim_{n\to \infty}\left(1+\frac{O\left(\frac{1}{n^4}\right)}{1-\frac1{2n^2}}\right)^{n^2}=\lim_{n\to \infty}\left(1+O\left(\frac{1}{n^4}\right)\right)^{n^2}=1$$

we arrive at the coveted limit

$$\lim_{n\to \infty}\cos^{n^2}(1/n)=e^{-1/2}$$

And we are done!

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$$\lim_{n\to\infty}\left(\cos\dfrac1n\right)^{n^2}=\lim_{n\to\infty}\left(1-\sin^2\dfrac1n\right)^{n^2/2}$$

$$=\left(\lim_{n\to\infty}\left(1-\sin^2\dfrac1n\right)^{-\frac1{\sin^2\frac1n}}\right)^{-\lim_{n\to\infty}\frac{n^2\sin^2\frac1n}2}$$

Set $-\sin^2\dfrac1n=\dfrac1m$

$$\lim_{n\to\infty}\left(1-\sin^2\dfrac1n\right)^{-\frac1{\sin^2\frac1n}}=\lim_{m\to\infty}\left(1+\dfrac1m\right)^m=?$$

For exponent set $\dfrac1n=h$ to get $$-\lim_{n\to\infty}\frac{n^2\sin^2\frac1n}2=-\dfrac12\left(\lim_{h\to0}\dfrac{\sin h}h\right)^2=?$$

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If $L=\lim_{x \to a} f (x)^{g (x)} $ assumes the form $1^{\infty} $ we can always write it as $L=e^{lim_{x\to a}g (x)(f (x)-1)} $. Now here let $\frac {1}{n}=u $ so as $n \infty ,u\to 0$. So we have $L=e^{\lim_{x \to 0}\frac{\cos(u)-1}{u^2}}=e^{\frac {-1}{2}} $ . The proof of the first statememt can be added if you want.

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$$\lim_{n \to \infty}\exp\underbrace{(n^2\ln(\cos(\frac{1}{n}))}_{a})\\$$exp is continous functio ,so find limit $a$ $$\lim_{n \to \infty}n^2\ln(\cos(\frac{1}{n})=\\ \lim_{n \to \infty}\frac{\ln(\cos(\frac{1}{n}))}{\frac{1}{n^2}}=\\hop\to \lim_{n \to \infty}\frac{\frac{\frac{-1}{n^2}\sin (\frac{1}{n})}{(\cos(\frac{1}{n})}}{\frac{-2}{n^3}}=\\ \lim_{n \to \infty}-\frac{\frac{\sin (\frac{1}{n})}{2\cos(\frac{1}{n})}}{\frac{1}{n}}=\\\frac{-1}{2}$$now $$a=\frac{-1}{2} \implies exp(a)=e^{\frac{-1}{2}}$$

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