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Let's consider $$a_n = n^2 \log \left(\cos \frac 1n\right)$$

It's easy to calculate $$\lim_{n\to\infty} a_n$$ by using l'Hospital/Taylor. But how to do it without using anything that involves derivatives? (Pure sequences!)

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    $\begingroup$ I tend to define the functions involved using derivatives, so that may be tricky ... $\endgroup$ – Hagen von Eitzen Jan 26 '16 at 20:06
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    $\begingroup$ This does converge: wolframalpha.com/input/?i=lim+n^2+log+%28cos+1%2Fn%29 $\endgroup$ – marmistrz Jan 26 '16 at 20:17
  • $\begingroup$ Why do you especially need to not to use L'Hospital's rule, Taylor's theorem or anything with derivatives? I think you asked a similar question before. $\endgroup$ – Arbuja Jan 26 '16 at 20:21
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    $\begingroup$ @Arbuja: Because it's fun? :) Because it's good to sometimes be able to solve stuff when things are not differentiable? $\endgroup$ – marmistrz Feb 4 '16 at 7:33
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I'll use two facts 1. $\lim_{x\to 0}\sin x/x = 1.$ 2. $\lim_{x\to 0}(1+ax +o(x))^{1/x} = e^a$ for any constant $a.$

From 1. we get, as $x\to 0,$

$$\frac{1-\cos x}{x^2} = \frac{1}{1+\cos x}\frac{1-\cos^2 x}{x^2} = \frac{1}{1+\cos x}\frac{\sin^2 x}{x^2} \to \frac{1}{2}\cdot 1^2 = \frac{1}{2}.$$

This shows $\cos x = 1 - (1/2)x^2 + o(x^2).$ Therefore

$$[\cos(1/n)]^{n^2} = [1+(-1/2)/n^2 + o(1/n^2)]^{n^2} \to e^{-1/2},$$

where we have used 2. above. Now apply $\ln$ to see the desired limit is $-1/2.$

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We have $$ \tag1n^2\ln\cos\frac 1n=\ln\left(\cos\frac 1n\right)^{n^2}$$ and $$\tag2\left(\cos\frac 1n\right)^{n^2}=\left(\sqrt{1-\sin^2\frac1n}\right)^{n^2}=\sqrt{\left(1-\sin^2\frac1n\right)^{n^2}}$$ For any $q<1$ we find $\epsilon>0$ such that $0<x<\epsilon$ implies $$\tag3qx<\sin x <x.$$ Hence for such $q$, for almost all $n$, $$\tag4\left(1-\frac1{n^2}\right)^{n^2}<\left(1-\sin^2\frac1n\right)^{n^2}<\left(1-\frac{q^2}{n^2}\right)^{n^2}$$ As $n\to\infty$, we have $\left(1-\frac1{n^2}\right)^{n^2}\to e^{-1}$ and $\left(1-\frac{q^2}{n^2}\right)^{n^2}\to e^{-q^2}$ (using $\lim_{n\to\infty}\left(1+\frac xn\right)^n$ as definition of $e^x$). Since $q<1$ was arbitrary, we conclude $\left(1-\sin^2\frac1n\right)^{n^2}\to e^{-1}$, hence $\left(\cos\frac 1n\right)^{n^2}\to e^{-\frac12}$ and ultimately $$\tag5n^2\ln\cos\frac 1n\to-\frac12.$$


To complete this without derivatives, we need an elementary proof of $(3)$. By geometric considerations about straight line lengths vs. arc lengths (or comparing suitable areas) one can show $\sin x<x<\tan x$ for $0<x<\frac\pi2$ and then $(3)$ follows from $\cos x\to 1$ as $x\to 0$.

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  • $\begingroup$ too good an answer +1. Only using basic inequalities and no derivatives as desired by OP. $\endgroup$ – Paramanand Singh Jan 28 '16 at 4:03
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In THIS ANSWER and THIS ONE, I showed using only (i) the limit definition of the exponential function and (ii) Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 1$$

for $x>0$.

Now, recall from basic geometry that the sine and cosine functions satisfy the inequalities

$$x\cos(x)\le \sin(x)\le x \tag 2$$

for $0\le x\le \pi/2$. From $(2)$, it is easy to show that the cosine function satisfies the inequalities

$$\sqrt{1-x^2}\le \cos(x)\le \frac{1}{\sqrt{1+x^2}} \tag 3$$

for $|x|<1$.

Then, putting $(1)$ and $(3)$ together (with $x=1/n<1$ ) reveals

$$n^2\left(\frac{\sqrt{1-(1/n)^2}-1}{\sqrt{1-(1/n)^2}}\right)\le n^2\log(\cos(1/n))\le n^2 \left(\frac{1-\sqrt{1+(1/n)^2}}{\sqrt{1+(1/n)^2}}\right) \tag 4$$

Rationalizing the numerators of the left-hand and right-hand sides of $(4)$, we obtain

$$\frac{-1}{\sqrt{1-(1/n)^2}(\sqrt{1-(1/n)^2}+1)}\le n^2\log(\cos(1/n)) \le \frac{-1}{\sqrt{1+(1/n)^2}(1+\sqrt{1+(1/n)^2})}$$

whereupon applying the Squeeze Theorem yields

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty }n^2\log(\cos(1/n))=-1/2}$$

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  • $\begingroup$ This an amazing answer because for extremely large n I still get a large negative number. $\endgroup$ – Arbuja Jan 26 '16 at 20:51
  • $\begingroup$ @Arbuja Thank you!! Much appreciate your comment. It made my day. ;-)) - Mark $\endgroup$ – Mark Viola Jan 26 '16 at 20:52
  • $\begingroup$ good use of the logarithmic inequality $(1)$. +1 $\endgroup$ – Paramanand Singh Jan 28 '16 at 4:04
  • $\begingroup$ @ParamanandSingh Thank you Paramanand! And nice to see you back my friend! - Mark $\endgroup$ – Mark Viola Jan 28 '16 at 5:53
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Write it as $$ \lim_{n\to\infty}\frac{\log(1-(1-\cos(1/n))}{1-\cos(1/n)} \frac{1-\cos(1/n)}{1/n^2} $$ By the basic limits $$ \lim_{t\to0}\frac{\log(1-t)}{t}=-1, \qquad \lim_{t\to0}\frac{1-\cos t}{t^2}=\frac{1}{2} $$ the limit is $-1/2$.

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