Limit abs value $\lim_{x\to \infty}{\left|\frac{x}{4}\right| \tan\left(\frac{2}{x}\right)}$ Is there any trick to evaluate this limit? I have no idea about next steps. Thanks
NO use L'Hopital, no derivate
$$\lim_{x\to \infty}{\left|\frac{x}{4}\right| \tan\left(\frac{2}{x}\right)}$$
 A: By setting $t=\frac{2}{x}$ we have
\begin{align}
\lim_{x\to\infty}\frac{x}{4}\tan\left(\frac{2}{x}\right)&=\lim_{t\to 0^+}\frac{\tan t}{2t}\\[4pt]
&=\frac{1}{2}\left(\lim_{t\to 0^+}\frac{1}{\cos t}\right)\left(\lim_{t\to 0^+}\frac{\sin t}{ t}\right)\\[4pt]
&=\frac{1}{2}\left(1\right)\left(1\right)\\[4pt]
&=\frac{1}{2}
\end{align}
A: Put $x = \frac{1}{u}$.
$$\lim_{u \to 0^+} \frac{\tan 2u}{4u} = \lim_{u \to 0^+} \frac{2\sec^2 2u}{4}$$
Using l'Hopital's rule.
A: Use equivalents: $\;\tan u\sim_0 u$. As$\;\dfrac2x\to 0$ if $x\to \infty$, we have
$$\frac{\lvert x\rvert}4\tan\frac2x\sim_\infty\frac{\lvert x\rvert}4\cdot\frac2x=(\operatorname{sgn}x)\frac12=\begin{cases}\dfrac12&\text{if}\enspace x\to +\infty,\\-\dfrac12&\text{if}\enspace x\to -\infty.\end{cases}$$
A: Notice, $$\lim_{x\to \infty}\left|\frac{x}{4}\right|\tan\left(\frac{2}{x}\right)$$
 $$=\lim_{x\to \infty}\left(\frac{x}{4}\right)\frac{\sin\left(\frac{2}{x}\right)}{\cos\left(\frac{2}{x}\right)}$$
$$=\frac{1}{2}\lim_{x\to \infty}\left(\frac{x}{2}\right)\sin\left(\frac{2}{x}\right)\cdot \lim_{x\to \infty}\frac{1}{\cos\left(\frac{2}{x}\right)}$$
$$=\frac{1}{2}\lim_{x\to \infty}\frac{\sin\left(\frac{2}{x}\right)}{\left(\frac{2}{x}\right)}\cdot \lim_{x\to \infty}\frac{1}{\cos\left(\frac{2}{x}\right)}$$
$$=\frac{1}{2}(1)\cdot (1)=\color{red}{\frac{1}{2}}$$
