1
$\begingroup$

Find the limit $$\lim_{x\to0}\left(\frac{1}{\arcsin x}-\frac{1}{\sin x}\right)$$

\begin{align} \lim_{x\to 0}\left(\frac{1}{\arcsin x}-\frac{1}{\sin x}\right)&=\lim_{x\to 0}\left(\frac{\sin x-\arcsin x}{\sin x\times \arcsin x}\right)\\ &=\lim_{x\to 0}\frac{x}{\arcsin x} \cdot\frac{x}{\sin x}\cdot \frac{\sin x-\arcsin x}{x^2} \end{align}

Here's where I'm stuck. I know from L'Hopital's Rule that $\frac{x}{\sin x}\to 1$, but what about $\frac{x}{\arcsin x}$? Or is there any other way than using L'Hopital to solve this question?

$\endgroup$
7
  • $\begingroup$ yes use lhopitals $\endgroup$
    – C Squared
    Sep 3, 2020 at 5:06
  • 4
    $\begingroup$ Even without L'Hopital $\frac{\sin x}{x}$ is a known limit. $\frac{x}{\arcsin x}$ is the same limit, just substitute $x = \sin\theta$ $\endgroup$
    – Ninad Munshi
    Sep 3, 2020 at 5:08
  • $\begingroup$ @NinadMunshi what do you mean by $x=sin\theta$? $\endgroup$
    – user815674
    Sep 3, 2020 at 5:11
  • 1
    $\begingroup$ Plug it in and see what happens. $\endgroup$
    – Ninad Munshi
    Sep 3, 2020 at 5:11
  • 1
    $\begingroup$ Oh, by $sin\theta\to0$ it implies that $\theta\to0$, correct? $\endgroup$
    – user815674
    Sep 3, 2020 at 5:16

3 Answers 3

Reset to default
2
$\begingroup$

Hint:

$$\dfrac{\sin x-\arcsin x}{\sin x\cdot\arcsin x}=x\cdot\dfrac{\dfrac{\sin x-x}{x^3}-\dfrac{\arcsin x-x}{x^3}}{\dfrac{\sin x}x\cdot\dfrac{\arcsin x}x}$$

Now use Are all limits solvable without L'Hôpital Rule or Series Expansion

$\endgroup$
3
  • $\begingroup$ Actually, you have to write $x^2$ instead of $x^3$. $\endgroup$
    – Angelo
    Sep 3, 2020 at 5:40
  • $\begingroup$ @Angelo no: in the numerator there is a sum, in the denominator a product $\endgroup$
    – Vincenzo Tibullo
    Sep 3, 2020 at 5:48
  • $\begingroup$ You are right, I am sorry. $\endgroup$
    – Angelo
    Sep 3, 2020 at 6:36
0
$\begingroup$

$$\frac{1}{\arcsin{x}}-\frac{1}{\sin{x}}=\frac{1}{x+\frac{x^3}{6}+...}-\frac{1}{x-\frac{x^3}{6}+...}\rightarrow0.$$

Your way also helps because $$\frac{\sin{x}-\arcsin{x}}{x^2}\rightarrow\frac{\cos{x}-\frac{1}{\sqrt{1-x^2}}}{2x}\rightarrow\frac{-\sin{x}-\frac{x}{\sqrt{(1-x^2)^3}}}{2}\rightarrow0.$$

$\endgroup$
0
$\begingroup$

$$L=\lim_{x\to 0} \left( \frac{1}{\sin^{-1} x}-\frac{1}{\sin x}\right)$$ Note that if $|x|$ is very small then $\sin^{-1} x=x+x^3/6+..., \sin x=x-x^3/6+... (1+x)^{k}=1+kx+...$ Then $$L=\lim_{x \to 0} \left( \frac{1}{x+x^3/6}-\frac{1}{x-x^3/6} \right)$$ $$\implies L=\lim_{x \to 0}\frac{1}{x}[(1+x^2/6)^{-1}-(1-x^2/6)^{-1}]=\lim_{x \to 0}\frac{1}{x}[(1-x^2/6)-(1+x^2/6)]=0$$

$\endgroup$

You must log in to answer this question.