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Solve the following equation:

\begin{eqnarray} \lim _{ x\to 0 } f(x)= \lim _{ x\to 0 } \frac { { \sin }^{ -1 }x }{ x }\\ \end{eqnarray}

The answer in my book is 1.

Can I do this?

\begin{eqnarray} \lim _{ x\to 0 } \frac { { \sin }^{ -1 }x }{ x }=\lim_{x\to 0} \frac{1}{x\sin x} \end{eqnarray}

I think $\sin^{-1}x$ is an inverse, not equal to $(\sin x)^{-1}$, so I can't do this. Right?

Are there any easily ways to deal with $\lim_{x\to a}$ including $\sin^{-1}x$, $\cos^{-1}x$ and $\tan^{-1}x$?

I have no idea to deal with this question.

Thank you for your attention.
haunted85
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Casper
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1 Answers1

6

HINT:

Put $\sin^{-1}x=h\implies x=\sin h$ and as $x\to0,h=\sin^{-1}x\to0$

lab bhattacharjee
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    Apart from Lab's correct answer, Casper, do you understand your mistake, i.e. that the inverse sine does NOT pop up as a regular sine in the denominator? That's very important! – imranfat Oct 14 '13 at 16:54
  • Since $h=sin^{−1}x$, then $x=sinh$. If x→0, h can be 0, $\pi$, or $2\pi$ [0, $2\pi$]. However, the denominator is still 0 so the calculator returned me "Error Math". Sorry, I don't understand.=( – Casper Oct 14 '13 at 17:06
  • @CasperLi, http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values – lab bhattacharjee Oct 14 '13 at 17:11
  • @Casper: There are a few problems with that. For one, the range of $\sin^{-1}x$ is $\left[-\frac\pi2,\frac\pi2\right],$ so $h$ cannot be $\pi$ or $2\pi$. For another, we aren't just "plugging in $0$," but rather taking a limit. – Cameron Buie Oct 14 '13 at 17:11