Calculating this limit without l'Hospital/Taylor/derivatives

Question

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!)

Answer 1

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 $$\tag5n^2\ln\cos\frac 1n\to-\frac12.$$

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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 $x\to 0$.

Answer 2

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

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

Answer 3

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 $-1/2.$

Answer 4

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 $-1/2$.