TITLE: Evaluate $\lim_{x\to \infty} \frac{\tan^{2}(\frac{1}{x})}{(\ln(1+\frac{4}{x}))^2}$
QUESTION [6 upvotes]: $$\lim_{x\to \infty} \frac{\tan^{2}(\frac{1}{x})}{(\ln(1+\frac{4}{x}))^2}$$
I came across this problem and I am having trouble evaluating it. I know that the whole limit will probably be $0$ and that both the numerator and denominator approach $0$.
How do I evaluate it? Using L'Hospital's rule leads to complex expressions, so I don't think that's a good method.
Thank you for the help.

REPLY [4 votes]: This is also pretty easy without using the taylor expansion. 
$$L=\lim_{x\to\infty}\frac{\tan^2(1/x)}{\ln^2(1+4/x)}=\left(\lim_{x\to\infty}\frac{\tan(1/x)}{\ln(1+4/x)}\right)^2$$
Let $u=1/x$.
$$\sqrt L=\lim_{u\to0}\frac{\tan(u)}{\ln(1+4u)}=\lim_{u\to0}\frac{\sin(u)}{\cos(u)\ln(1+4u)}$$
$$=\lim_{u\to0}\frac{\sin(u)}{\ln(1+4u)}$$
Now we can apply L'Hospital.
$$\sqrt L=\lim_{u\to 0}\frac{1}{4}\cos(u)(4u+1)=\frac 1 4$$
$$\lim_{x\to\infty}\frac{\tan^2(1/x)}{\ln^2(1+4/x)}=\frac{1}{16}$$