I have to evaluate the following without L'Hopital's rule
$$\lim_{x\to\infty} x\tan(1/x)$$
I can simplify this to be $$\lim_{x\to\infty} x\sin(1/x)$$
because $$\lim_{x\to\infty} \cos(1/x) = 1$$
However, after that, I'm totally lost. L'Hopital's rule seems like my only option. Can someone help me out?
If you are allowed to use the fact that $\lim_{t\to0}\frac{\sin(t)}{t}=1$, then setting $t=\frac{1}{x}$ yields $$ \lim_{x\to\infty}x\sin\Big(\frac{1}{x}\Big)=\lim_{x\to\infty}\frac{\sin(\frac{1}{x})}{\frac{1}{x}}=\lim_{t\to 0}\frac{\sin(t)}{t}=1$$
After a change of variables, this is just
$$\lim_{t \to 0^+} \frac 1 t \tan t$$
which you can evaluate via the definition of the derivative of tangent, or geometrically prove that $\lim_{t \to 0} \frac 1 t \sin t = 1$.
$$\lim\limits_{x\to\infty} x\tan\frac{1}{x}$$ Let $t=\frac{1}{x}$, then $$\lim\limits_{t\to 0} \frac{\tan t}{t}=\lim\limits_{t\to 0} \frac{\sin t}{t\cos t}$$ $$=\left(\lim\limits_{t\to 0} \frac{\sin t}{t}\right)\left(\lim\limits_{t\to 0}\frac{1}{\cos t}\right)=1$$
