# 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 $$\frac{1}{1}*\frac{3}{2} = \frac{3}{2}$$

Your approach is correct assuming you already know $\lim \frac{\sin x}{x}=1$. Make sure you justify your steps: you can only separate the big limit in a product/quotient of separate limits assuming those exist (which you prove at the end), and you can say that there exists a $y\in(0,\frac \pi2)$ such that $\tan y = \frac 3x$ by bijectivity of $\tan:(0,\frac \pi 2)\to (0,\infty)$.
Since $\arctan{}'(x) = 1/(1 + x^2)$, $arctan'(0) = 1$. Therefore you have $$\lim_{x\to 0} {\arctan(x)\over x} = 1.$$ This will enable you to do the rest.
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.