Limit with abs $\lim_{x\to0^-}{\frac{|\tan(-2x)|}{\arcsin(-5x)}}$

Question

It there any rule about abs limits? Thanks NO use L'Hopital, no derivate

$$\lim_{x\to0^-}{\frac{|\tan(-2x)|}{\arcsin(-5x)}}$$

Answer 1

By definition $$|x|=\left\{\begin{array}{cc} x,& x\geq 0\\ -x,& x<0\end{array} \right.$$ When we use the limit we have to consider the left and right limit of the point(s) in which the abs. value is zero, in this case it is for $x=0$, and we have that $$\lim_{x\rightarrow 0-}|x|=\lim_{x\rightarrow 0-} (-x) \; \; \text{and} \; \;\lim_{x\rightarrow 0+}|x|=\lim_{x\rightarrow 0+} x$$

For your problem, you get. $$\lim_{x\rightarrow 0-} \frac{|\tan (-2x)|}{\arcsin (-5x)}=\lim_{x\rightarrow 0-} \frac{-\tan (-2x)}{\arcsin (-5x)}=\lim_{x\rightarrow 0-} \frac{\tan (2x)}{\arcsin (-5x)}=\left\{\begin{array}{c} x=0-\varepsilon\\ \varepsilon >0 \\ \varepsilon \rightarrow 0\end{array} \right\}=\lim_{\varepsilon \rightarrow 0} \frac{\tan (-2\varepsilon)}{\arcsin (5\varepsilon)}$$ $$=-\lim_{\varepsilon \rightarrow 0} \frac{\tan (2\varepsilon)}{\arcsin (5\varepsilon)}=-\lim_{\varepsilon \rightarrow 0} \frac{\sin(2\varepsilon)}{\cos (2\varepsilon) \arcsin (5\varepsilon)}=\left\{\begin{array}{c} \sin x\sim x,\; \arcsin x\sim x\\x \rightarrow 0\end{array} \right\}=-\lim_{\varepsilon \rightarrow 0} \frac{2\varepsilon}{\cos (2\varepsilon) 5\varepsilon}=-\frac{2}{5}$$

Answer 2

For $x\to0,x<0$, $\tan(-2x)>0$ hence $\frac{|\tan(-2x)|}{\arcsin(-5x)}=\frac{\tan(2x)}{\arcsin(5x)}=\frac{\sin(2x)}{\arcsin(5x)}\frac{1}{\cos(2x)}$

Thus $\lim_{x\to0^-}{\frac{|\tan(-2x)|}{\arcsin(-5x)}}=\lim_{x\to0^-}\frac{\sin(2x)}{\arcsin(5x)}\frac{1}{\cos(0)}$

By l'Hopital (PS : I forgot you didn't want to use l'Hopital's rule, I'll edit this - Edit : see zemelovac's answer, there's no need for the rule to solve the following limit) , $\lim_{x\to0}\frac{\sin(2x)}{\arcsin(5x)}=\lim_{x\to0}\frac{2}5\sqrt{1-25x^2}\cos(2x)=\frac{2}5$

Thus $\boxed{\lim_{x\to0^-}{\frac{|\tan(-2x)|}{\arcsin(-5x)}}=\frac{2}5}$