Limit and L'hopital's Rule

Question

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?

score 1 · Answer 1 · answered Jan 18 '19 at 03:03

1

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

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

answered Jan 18 '19 at 03:03

Tsemo Aristide

87,475

It doesn't exist https://www.wolframalpha.com/input/?i=lim_(x-%3E0)+(x+sin%5E(-1)(x))%2F(x+-+sin(x)) – AfronPie Jan 18 '19 at 03:07
Your mathjax could use some help. – zhw. Jan 18 '19 at 20:03

score 0 · Accepted Answer · answered Jan 18 '19 at 05:51

0

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

The first & the last limits are elementary and

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

answered Jan 18 '19 at 05:51

lab bhattacharjee

274,582

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 '19 at 15:08
@Ashish, It held if at least one of them is non-zero finite – lab bhattacharjee Jan 19 '19 at 14:23

Limit and L'hopital's Rule

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