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.

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!
