0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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

The first & the last limits are elementary and

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Ashish Jan 18 at 15:08
  • $\begingroup$ @Ashish, It held if at least one of them is non-zero finite $\endgroup$ – lab bhattacharjee Jan 19 at 14:23
1
$\begingroup$

$sin^{-1}(x)= x+{x^3\over 6}+O(x^3)$ 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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.