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How to calculate $\lim\limits_{n\to \infty} \left( \cos(1/n)-\sin(1/n) \right) ^n $?

Since $\lim\limits_{n\to \infty} \frac {\cos(1/n)-\sin(1/n) }{1-1/n} = 1 $, I guess that the limit above is $\frac{1}{e}$, but since the form $(\to 1)^{\to \infty}$ is indeterminate, I don't know how to prove it formally.

Thanks!

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  • $\begingroup$ You are right as $\cos1/n-\sin1/n=1-1/n+o(1/n)$. $\endgroup$ – Yves Daoust Jan 30 at 13:17
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Use $$\log\left[\cos\left(\frac{1}{n}\right)-\sin\left(\frac{1}{n}\right)\right]^n = n\log\left[\cos\left(\frac{1}{n}\right)-\sin\left(\frac{1}{n}\right)\right] = \frac{\log\left[\cos\left(\frac{1}{n}\right)-\sin\left(\frac{1}{n}\right)\right]}{\frac{1}{n}}$$ and then apply l'Hospital's rule.

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  • $\begingroup$ You must first prove that the inner limit is not zero - otherwise the function is undefined at infinity. $\endgroup$ – Lucas Henrique Jan 30 at 13:57
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You can get rid of the trigonometric functions by rewriting the exponent $n$ using

$$\frac1n=\dfrac{\dfrac1n}{\sin\dfrac1n}\sin\dfrac1n.$$

As the larger fraction tends to $1$, it can be ignored.

Now, with $t:=\sin\dfrac1n$, we have

$$\lim_{t\to0}\left(\sqrt{1-t^2}-t\right)^{1/t}.$$

Using Taylor,

$$\sqrt{1-t^2}-t=1-t+o(t)$$ and the limit is the same as

$$\lim_{t\to0}\left(1-t\right)^{1/t}.$$


Or by L'Hospital on the logarithm,

$$\frac{\log(\sqrt{1-t^2}-t)}t\to\frac{-\dfrac t{\sqrt{1-t^2}}-1}{\sqrt{1-t^2}-t}\to-1.$$

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Since we recall that $$\sin x=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots $$ and $$\cos x=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\cdots$$ for all $x$, then $$\cos\left(\frac1n\right)-\sin\left(\frac1n\right)=1-\frac1n+\dots$$ and so $$\lim\limits_{n\to \infty} \left(\cos\left(\frac1n\right)-\sin\left(\frac1n\right) \right) ^n=\lim\limits_{n\to \infty} \left(1- \frac1n \right) ^n=e^{-1}\, .$$

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  • $\begingroup$ Thanks a lot!!! $\endgroup$ – Dr. John Jan 31 at 20:13
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We have : $\lim\limits_{n\to \infty} \left( \cos(\frac{1}{n})-\sin(\frac{1}{n}) \right) ^n$$=\lim\limits_{n\to \infty}(1+\cos(\frac{1}{n})-\sin(\frac{1}{n})-1)^n$$=\lim\limits_{n\to \infty}\left[(1+\cos(\frac{1}{n})-\sin(\frac{1}{n})-1)^{\frac{1}{\cos(\frac{1}{n})-\sin(\frac{1}{n})=1}}\right]^{n(\cos(\frac{1}{n})-\sin(\frac{1}{n})-1)}$$=e^{\lim\limits_{n\to \infty}n(\cos(\frac{1}{n})-\sin(\frac{1}{n})-1)}$ But $\lim\limits_{n\to \infty}n(\cos(\frac{1}{n})-\sin(\frac{1}{n})-1)$$=\lim\limits_{n\to \infty}\frac{\cos(\frac{1}{n})-\sin(\frac{1}{n})-1}{\frac{1}{n}}$$=\lim\limits_{x\to 0}\frac{\cos x-\sin x-1}{x}$$=$$\lim\limits_{x\to 0}(-\sin x-\cos x)=-1$,so your limit equals $\frac{1}{e}$.

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You may consider

  • $\left(\frac{\cos x - \sin x}{1-x}\right)^{\frac{1}{x}}$ for $x\to 0$ and use
  • $\lim_{t\to 0} (1+t)^{\frac{1}{t}}= e$ \begin{eqnarray*} \left(\frac{\cos x - \sin x}{1-x}\right)^{\frac{1}{x}} & = & \left( \underbrace{\left(1 + \frac{\cos x - \sin x -1 +x}{1-x}\right)^{\frac{1-x}{\cos x - \sin x -1 +x}}}_{\stackrel{x\to 0}{\longrightarrow}e}\right)^{\underbrace{\frac{\cos x - \sin x -1 +x}{x-x^2}}_{\stackrel{L'Hop}{\sim}\frac{-\sin x - \cos x +1}{1-2x}\stackrel{x\to 0}{\longrightarrow}0}}\\ & \stackrel{x\to 0}{\longrightarrow} & e^0 = 1 \end{eqnarray*}
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Using the well known limit $$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n=e\tag{1}$$ it is easy to prove that $$\lim_{n\to\infty} \left(1-\frac{1}{n}\right)^n=\lim_{n\to\infty}\dfrac{1}{\left(1+\dfrac{1}{n-1}\right)^{n-1}}\cdot\frac{n-1}{n}=\frac{1}{e}\tag{2}$$ Next we use the following lemma :

Lemma: If $\{a_n\} $ is sequence such that $n(a_n-1)\to 0$ then $a_n^n\to 1$.

We can now choose $$a_n=\dfrac{\cos\left(\dfrac{1}{n}\right)-\sin\left(\dfrac{1}{n}\right)} {1-\dfrac{1}{n}} $$ and note that $$n(a_n-1)=n\left(\dfrac{n\cos\left(\dfrac{1}{n}\right)-n\sin\left(\dfrac{1}{n}\right)-n+1} {n-1}\right) $$ and the above clearly tends to $0$ so that by lemma above $a_n^n\to 1$ and therefore using $(2)$ the desired limit is $1/e$.

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