Limit of function $\lim\limits_{x \to - \infty } {{\arcsin (-{5 \over x})} \over {|\tan ({2 \over x})|}}$

Question

I am evaluating limits of function. I dont know how to start evaluate this limit, I cannot use L Hopital's rule.

Thankx.

$$\lim\limits_{x \to - \infty }{{\arcsin (-{5 \over x})} \over {|\tan ({2 \over x})|}}$$

Answer 1

Hint: As $u\to 0,$ both $(\arcsin u) /u, (\tan u)/u \to 1,$ because these terms $ \to \arcsin'(0), \tan'(0)$ respectively, by the definition of the derivative.

Answer 2

Using the fact that, for $t \to 0$, $$\begin{align} \arcsin t &= t + o(t^2)\\ \tan t &= t + o(t^2) \end{align}$$

we get, for $x \to -\infty$, $$\begin{align} \arcsin\left(-\frac5x\right) &= \frac5x + o(x^{-2})\\ \tan \frac2x &= \frac2x + o(x^{-2}). \end{align}$$

Hence, it follows that $$\lim_{x \to -\infty }\frac{\arcsin\left(-\frac5x\right)}{\left|\tan \frac2x\right|} = \lim_{x \to -\infty} \frac{\frac5x + o(x^{-2})}{\frac2x + o(x^{-2})} = \frac52$$

Answer 3

First change of the variable $t = 1/x$ so that the limit process is as $t \to 0$-. For sufficiently small $t$, then let $u = \arcsin(-5t)$ (which is monotone transformation and $u \to 0$+) so that $-5t = \sin u$, hence $$\tan(2t) = \frac{\sin(2t)}{\cos(2t)} = \frac{\sin\left(-\frac{2}{5}\sin u\right)}{\cos\left(-\frac{2}{5}\sin u\right)}$$ and the required limit becomes $$\lim_{u \to 0+} \frac{u|\cos\left(-\frac{2}{5}\sin u\right)|}{|\sin\left(-\frac{2}{5}\sin u\right)|} = -\lim_{u \to 0+} \frac{u\cos\left(-\frac{2}{5}\sin u\right)}{\sin\left(-\frac{2}{5}\sin u\right)} = \frac{5}{2}\lim_{u \to 0+} \frac{u}{\sin u} \times \cos\left(-\frac{2}{5}\sin u\right) = \frac{5}{2}.$$ where we used the relation $y \sim \sin y$ as $y \to 0$ to replace $\sin\left(-\frac{2}{5}\sin u\right)$ by $-\frac{2}{5}\sin u$.