How to evaluate $\lim_{x\to\infty}\arctan (4/x)/ |\arcsin (-3/x)|$?

Question

I don't know how to start evaluate this limit, I cannot use L Hopital's rule. Thank you very much for all responses.

$$\mathop {\lim }\limits_{x \to + \infty } \left({{\arctan \left({4 \over x}\right)} \over {\left|\arcsin \left( - {3 \over x}\right)\right|}}\right)$$

Answer 1

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

$$\eqalign{ & \,\,\,\,\mathop {\lim }\limits_{x \to + \infty } \frac{{\arctan (\frac{4}{x})}}{{|\arcsin ( - \frac{3}{x})|}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\arctan (\frac{4}{x})}}{{| - \arcsin (\frac{3}{x})|}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\arctan (\frac{4}{x})}}{{|\arcsin (\frac{3}{x})|}} \cr & = \mathop {\lim }\limits_{x \to + \infty } \frac{{\arctan (\frac{4}{x})}}{{\arcsin (\frac{3}{x})}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{4\frac{{\arctan (\frac{4}{x})}}{{\frac{4}{x}}}}}{{3\frac{{\arcsin (\frac{3}{x})}}{{\frac{3}{x}}}}} = \frac{{4\mathop {\lim }\limits_{x \to + \infty } \frac{{\arctan (\frac{4}{x})}}{{\frac{4}{x}}}}}{{3\mathop {\lim }\limits_{x \to + \infty } \frac{{\arcsin (\frac{3}{x})}}{{\frac{3}{x}}}}} = \frac{4}{3} \cr} $$

Answer 2

Using L'Hospital rule( $\arcsin ( - {3 \over x})<0$ at infinity )

$$\mathop {\lim }\limits_{x \to + \infty } {{\arctan ({4 \over x})} \over {|\arcsin ( - {3 \over x})|}}= -\mathop {\lim }\limits_{x \to + \infty } {{\arctan ({4 \over x})} \over {\arcsin ( - {3 \over x})}}= -\mathop {\lim }\limits_{x \to + \infty } \frac{-\frac{4}{x^2+16}} {\frac{3}{\sqrt{1-\frac{9}{x^2}} x^2}}= \mathop {\lim }\limits_{x \to + \infty }\frac{4 \sqrt{1-\frac{9}{x^2}}}{3 \left(\frac{16}{x^2}+1\right)}= \frac{4}3 $$