# Find $\lim_{x\to \infty}\left(x^2\ln\left(\cos \frac{4}{x}\right)\right)$

How do you evaluate $$\lim_{x\to\infty}\left(x^2\ln\left(\cos\frac{4}{x}\right)\right)$$

I know that you have to rewrite the expression as a quotient and use L'Hospital's Rule but I cant seem to figure it out.

• Let $1/x=t$, or slightly better, $4/x=t$. – André Nicolas Oct 11 '15 at 23:26

Use L'Hopital twice. The limit is:

$$\lim_{x\to +\infty}\frac{\ln\left(\cos\left(\frac{4}{x}\right)\right)}{\frac{1}{x^2}}=\lim_{x\to +\infty}\frac{\frac{1}{\cos\left(\frac{4}{x}\right)}\left(-\sin\left(\frac{4}{x}\right)\right) \frac{-4}{x^2}}{\frac{-2}{x^3}}$$

$$=-2\lim_{x\to +\infty}\frac{\tan\left(\frac{4}{x}\right)}{\frac{1}{x}}=-2\lim_{x\to +\infty}\frac{\frac{1}{\cos^2 \left(\frac{4}{x}\right)}\frac{-4}{x^2}}{\frac{-1}{x^2}}$$

$$=-8\lim_{x\to +\infty}\frac{1}{\cos^2\left(\frac{4}{x}\right)}=-8\cdot \frac{1}{1^2}=-8$$

In this approach, we use some standard inequalities along with the squeeze theorem. First, recall that the log function satisfies the inequalities

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

for $x>0$. Then using $(1)$, it is straightforward to see that

$$x^2\frac{\left(\cos\left(\frac 4x\right)-1 \right)}{\cos\left(\frac 4x\right)}\le x^2\log \cos\left(\frac 4x\right)\le x^2\left(\cos\left(\frac 4x\right)-1 \right) \tag 2$$

Now, using the trigonometric identity $\cos x-1=-2\sin^2(x/2)$ in $(2)$ yields

$$-2x^2\left(\frac{\sin^2\left(\frac 2x\right)}{\cos\left(\frac 4x\right)}\right)\le x^2\log \cos\left(\frac 4x\right)\le -2x^2\sin^2\left(\frac 2x\right) \tag 3$$

Next, recall that the sine function satisfies the inequalities

$$x\cos x\le \sin x\le x \tag 4$$

for $x\ge 0$. So, using $(4)$ in $(3)$ yields

$$-8\left(\frac{\cos^2\left(\frac2x\right)}{\cos\left(\frac 4x\right)}\right)\le x^2\log \cos\left(\frac 4x\right)\le -8 \tag 5$$

Finally, we have from the squeeze theorem

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty} x^2\log \cos\left(\frac 4x\right)=-8}$$

and we are done!

Putting $x = 1/t$ we see that as $x \to \infty, t \to 0^{+}$. Hence \begin{align} L &= \lim_{x \to \infty}x^{2}\log\cos\left(\frac{4}{x}\right)\notag\\ &= \lim_{t \to 0^{+}}\frac{\log\cos 4t}{t^{2}}\notag\\ &= \lim_{t \to 0^{+}}\frac{\log(1 - 2\sin^{2} 2t)}{t^{2}}\notag\\ &= \lim_{t \to 0^{+}}\frac{\log(1 - 2\sin^{2} 2t)}{2\sin^{2}2t}\cdot\frac{2\sin^{2}2t}{(2t)^{2}}\cdot 4\notag\\ &= -1\cdot 2\cdot 4 = -8\notag \end{align}