Do I have to use L'hoptial's rule? How can this be used in this instance?


we have $$\lim_{x\to0}\frac{\arcsin x}{x}$$ making $u=\arcsin x,x=\sin u,x\mapsto0= u\mapsto0$ $$\lim_{x\to0}\frac{\arcsin x}{x}=\lim_{u\to 0}\frac{u}{\sin u}=\lim_{u\to 0}\frac{1}{\frac{\sin u}{u}}=\frac{1}{\displaystyle\lim_{u\to0}\frac{\sin u}{u}}=1$$ you can use l'hospital since its a limit of type 0/0 wich gives $$\lim_{x\to 0}\frac{\arcsin x}{x}= \lim_{x\to 0}\frac{1}{\sqrt{1-x^2}}=1$$

  • $\begingroup$ What if it was $x \to \infty$? $\endgroup$ – RobChem Dec 12 '14 at 11:33
  • $\begingroup$ @RobChem I believe arcsin may is only defined on [-1,1],so allowing $x\rightarrow\infty$ is incorrect $\endgroup$ – Tony Ma Jan 31 '18 at 12:15

Let $x = \sin y$. Now, $x \to 0 \Rightarrow y \to 0$, giving us the following limit:

$$ \lim\limits_{y \to 0} \frac{\arcsin (\sin y)}{\sin y} = \lim\limits_{y \to 0} \frac{y}{\sin y} = \frac{1}{\lim\limits_{y \to 0} \frac{\sin y}{y}} = 1 $$

  • $\begingroup$ What if it was $x \to \infty$? $\endgroup$ – RobChem Dec 12 '14 at 11:32

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