# 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

$$\lim_{x\to \infty}{\left|\frac{x}{4}\right| \tan\left(\frac{2}{x}\right)}$$

• Hint: ${\tan(2/x)\over(2/x)}\to1$. – Aretino Nov 2 '15 at 19:21

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}

• I can't resist to give +1! (10k) – Olivier Oloa Nov 2 '15 at 19:35
• @OlivierOloa Thak you very much :). – Ángel Mario Gallegos Nov 2 '15 at 19:54
• thanks, why you change this $|\frac{x}{4}|$ to $\frac{x}{4}$ ?? – DavidM Nov 4 '15 at 18:33
• @David as $x\to \infty$ we have $x>0$, then $|\frac{x}{4}|=\frac{x}{4}$ – Ángel Mario Gallegos Nov 4 '15 at 19:01

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.

• thanks, but i cannot use L'hopital – DavidM Nov 2 '15 at 19:25
• No sweat. $\frac{1}{\cos 2u} \to 1$, and $\frac{\sin 2u}{4u} \to \frac{1}{2}$. SInce both limits exist, the limit of the product equals the product of the limits. – stochasticboy321 Nov 2 '15 at 19:28
• The answer deserves an upvote. +1 – Olivier Oloa Nov 2 '15 at 19:39

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 general answer: +1. – Olivier Oloa Nov 2 '15 at 19:38

Notice, $$\lim_{x\to \infty}\left|\frac{x}{4}\right|\tan\left(\frac{2}{x}\right)$$ $$=\lim_{x\to \infty}\left(\frac{x}{4}\right)\frac{\sin\left(\frac{2}{x}\right)}{\cos\left(\frac{2}{x}\right)}$$ $$=\frac{1}{2}\lim_{x\to \infty}\left(\frac{x}{2}\right)\sin\left(\frac{2}{x}\right)\cdot \lim_{x\to \infty}\frac{1}{\cos\left(\frac{2}{x}\right)}$$ $$=\frac{1}{2}\lim_{x\to \infty}\frac{\sin\left(\frac{2}{x}\right)}{\left(\frac{2}{x}\right)}\cdot \lim_{x\to \infty}\frac{1}{\cos\left(\frac{2}{x}\right)}$$ $$=\frac{1}{2}(1)\cdot (1)=\color{red}{\frac{1}{2}}$$