0

I am trying to show that $$\lim_{x\to 0}\dfrac{\arcsin x}{x}=1$$ I am new to inverse trigonometric functions, so I am sorry if it's obvious.

So if we put $\arcsin x=t,$ then $\sin t=x$. How do I say where t goes when $x\to 0$?

Trifon
  • 113
  • Are you familiar with the limit involving $\frac {\sin x}x$ where $x\to 0$? – abiessu Nov 12 '23 at 22:21
  • 1
    @abiessu, yes, I'm. After we change the variable we have that the givin limit is equal to $\lim_{t\to}\dfrac{t}{\sin t}$, but I am not sure whut number $t$ approaches in the new limit as $x\to 0$. That's my question. – Trifon Nov 12 '23 at 22:22
  • 1
    Take it in reverse; if $t\to 0$ then what value does $\sin t$ go to? – abiessu Nov 12 '23 at 22:24
  • to $0.$........................................ – Trifon Nov 12 '23 at 22:25
  • The typical assumption is that $\sin^{-1}x$ gives values in the interval $[-\frac {\pi}2,\frac {\pi}2]$. Otherwise the original limit could not exist. – abiessu Nov 12 '23 at 22:27
  • 1
    To understand why $t\to\arcsin0$, it's a worthwhile exercise to do this through the $\varepsilon$-$\delta$ definition of a limit, by relating it to the same definition of $\lim_{t\to0}\frac{t}{\sin t}$. – J.G. Nov 12 '23 at 22:34
  • Alternative approach is to obtain the Taylor series for the arcsin function. See, for example, the answer of noahklein in this MathSE article. – user2661923 Nov 13 '23 at 01:02
  • A less fiddly version of @user2661923's suggestion is L'Hôpital's rule. – J.G. Nov 13 '23 at 08:19

0 Answers0