# Help with inverse trig derivatives with L'Hopital rule

so I have a problem with my teacher's notes and they have confused me. take this equation \begin{align} \lim _{x\to0} \frac{\arcsin x}{ \sin x} \end{align}

if we do the derivative of the top and bottom using l'hopitals rule we get

\begin{align} (\arcsin x)'= \frac{1}{ \sqrt{1-x^2}} \end{align} \begin{align} (\sin x)'= \cos x \end{align}

but when my teacher put them together she did't explain how she somehow got this. \begin{align} \lim _{x\to0} \frac{\sqrt{1-x^2}}{ \cos x} \end{align}

• It's a typo, basically – b00n heT Apr 28 '17 at 18:25
• okay so she didn't do some wizard magic to make the fraction disappear. Then would the problem still make sense if the numerator was a fraction? – M.Bucciacchio Apr 28 '17 at 18:28
• Yep... your teacher is wrong. – Jed Apr 28 '17 at 18:29
• thank goodness. thank you guys very much. – M.Bucciacchio Apr 28 '17 at 18:31
• There is no point in using align when you are writing a single equation. – A---B Apr 28 '17 at 18:41

HINT:

Note that

$$\frac{\arcsin(x)}{\sin(x)}=\frac{\arcsin(x)}{x}\,\frac{x}{\sin(x)}$$

$$\frac{\arcsin(x)'}{\sin(x)'} = \frac{\frac{1}{\sqrt{1 -x^2}}}{\cos(x)} \cdot \frac{\sqrt{1 -x^2}}{\sqrt{1 -x^2}} = \frac{1}{\sqrt{1 -x^2}\cos(x)}$$
$$\lim_{x\to 0} \frac{1}{\sqrt{1 -x^2}\cos(x)}$$