<s>[INST] How to solve this limit without using L'Hospital's Rule? $$ 
\lim\limits_{x\to{\infty}}\frac{\arctan \frac{3}{x}}{|\arctan \frac{2}{x}|}
$$
Can anybody help me to solve this one ?
I ve done somethig like this but im not sure if it is the correct aproach.
$$ 
\lim\limits_{x\to{\infty}}\frac{\arctan\frac{3}{x}}{|\arctan \frac{2}{x}|}  =  
\lim\limits_{x\to{\infty}}\frac{\arctan \frac{3}{x}}{\arctan \frac{2}{x}} = 
\lim\limits_{x\to{\infty}} \frac{\arctan \frac{3}{x}}{\arctan \frac{2}{x}} * \frac{\frac{\frac{3}{x}}{\frac{3}{x}}}{\frac{\frac{2}{x}}{\frac{2}{x}}} = \frac{\lim\limits_{x\to{\infty}}\frac{\arctan \frac{3}{x}}{\frac{3}{x}}}{\lim\limits_{x\to{\infty}}\frac{\arctan \frac{2}{x}}{\frac{2}{x}}}*\lim\limits_{x\to{\infty}}\frac{\frac{3}{x}}{\frac{2}{x}}
$$
and then for each limit with arctan i ve substitued 3/x and 2/x by tan y and tan z
$$
\frac{\lim\limits_{y\to{0^+}}\frac{\arctan \tan y}{\tan y}}{\lim\limits_{z\to{0^+}}\frac{\arctan \tan z}{\tan z}}*\lim\limits_{x\to{\infty}}\frac{\frac{3}{x}}{\frac{2}{x}}
$$
and then for each limit which goes to 0^+ i ve done this
$$
\lim\limits_{y\to{0^+}}\frac{\arctan \tan y}{\tan y} = \lim\limits_{y\to{0^+}}\frac{y}{\tan y} = \lim\limits_{y\to{0^+}}\frac{y}{\frac{\sin y}{\cos y}} = 1
$$
so in the end 
$$
\frac{1}{1}*\frac{3}{2} = \frac{3}{2}
$$
 [/INST] Let $u := \frac{1}{x}$, then $$\lim_{u\to 0}\frac{\arctan 3u}{\arctan 2u} = \frac{3}{2}$$ because $\arctan u\sim u$, when $u\to 0$. As $x\to\infty$, then $|\arctan x|\to \frac{\pi}{2}$: absolute value signs unnecessary.
</s>