# Limit and L'hopital's Rule

How do you evaluate the following limit? $$\lim_{x \to 0} \dfrac{x\sin^{-1}x}{x-\sin{x}}$$ When I is L'Hopital's rule twice, I get: $$\lim_{x \to 0} \dfrac{(x^2+2)\csc x}{(1-x^2)^{3/2}}$$ Which doesn't exits. If the limit DNE then can't use L'Hopital's rule.

So, how do I find this limit?

$$\lim_{x\to0}\dfrac{x\sin^{-1}x}{x-\sin x}=\lim_{x\to0}\dfrac{\sin^{-1}x}x\cdot\lim_{x\to0}\dfrac1{\dfrac{x-\sin x}{x^3}}\cdot\lim_{x\to0}\dfrac1x$$

The first & the last limits are elementary and

for the second use Are all limits solvable without L'Hôpital Rule or Series Expansion

• I was under the impression that the limit property lim[f(x)g(x)]=lim[f(x)]lim[g(x)] held only if the limits were finite not for DNE. – Ashish Jan 18 at 15:08
• @Ashish, It held if at least one of them is non-zero finite – lab bhattacharjee Jan 19 at 14:23

$$sin^{-1}(x)= x+{x^3\over 6}+O(x^3)$$ implies that $$xsin^{-1}(x)=x^2+{x^4\over 6}+O(x^4)$$,

$$sin(x)=x-{x^3\over 6}+O(x^3)$$ implies that $$x-sin(x)={x^3\over 6}+O(x^3)$$ implies that the limit is

$$lim_{x\rightarrow 0}{{x^2+{x^4\over 6}+O(x^4)}\over{{x^3\over 6}+O(x^3)}}=+\infty.$$