How to evaluate $\lim_{x\to\infty}\arctan (4/x)/ |\arcsin (-3/x)|$? I don't know how to start evaluate this limit, I cannot use L Hopital's rule.
Thank you very much for all responses.
$$\mathop {\lim }\limits_{x \to  + \infty } \left({{\arctan \left({4 \over x}\right)} \over {\left|\arcsin \left( - {3 \over x}\right)\right|}}\right)$$
 A: I hope that you know the limits
$$\mathop {\lim }\limits_{t \to {0^ + }} \frac{{\arctan (t)}}{t} = \mathop {\lim }\limits_{t \to {0^ + }} \frac{{\arcsin (t)}}{t} = 1$$
the you can write
$$\eqalign{
  & \,\,\,\,\mathop {\lim }\limits_{x \to  + \infty } \frac{{\arctan (\frac{4}{x})}}{{|\arcsin ( - \frac{3}{x})|}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\arctan (\frac{4}{x})}}{{| - \arcsin (\frac{3}{x})|}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\arctan (\frac{4}{x})}}{{|\arcsin (\frac{3}{x})|}}  \cr 
  &  = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\arctan (\frac{4}{x})}}{{\arcsin (\frac{3}{x})}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{4\frac{{\arctan (\frac{4}{x})}}{{\frac{4}{x}}}}}{{3\frac{{\arcsin (\frac{3}{x})}}{{\frac{3}{x}}}}} = \frac{{4\mathop {\lim }\limits_{x \to  + \infty } \frac{{\arctan (\frac{4}{x})}}{{\frac{4}{x}}}}}{{3\mathop {\lim }\limits_{x \to  + \infty } \frac{{\arcsin (\frac{3}{x})}}{{\frac{3}{x}}}}} = \frac{4}{3} \cr} $$
A: Using L'Hospital rule( $\arcsin ( - {3 \over x})<0$ at infinity )
$$\mathop {\lim }\limits_{x \to  + \infty } {{\arctan ({4 \over x})} \over {|\arcsin ( - {3 \over x})|}}=
-\mathop {\lim }\limits_{x \to  + \infty } {{\arctan ({4 \over x})} \over {\arcsin ( - {3 \over x})}}=
-\mathop {\lim }\limits_{x \to  + \infty } \frac{-\frac{4}{x^2+16}} {\frac{3}{\sqrt{1-\frac{9}{x^2}} x^2}}=
\mathop {\lim }\limits_{x \to  + \infty }\frac{4 \sqrt{1-\frac{9}{x^2}}}{3 \left(\frac{16}{x^2}+1\right)}=
\frac{4}3
$$
