Calculating the Limit Using L'Hopital's Rule and Exponential Properties

[gipc]

I know I should apply L'Hopital's rule and use a^b=e^(b*ln(a)) but I can't finish the calculations.

limit as x->0 ((arcsin(x))/x) ^(1/x^2)

 

[lanedance]

thats a tricky one, so going with what you said
\lim_{x \to 0} (\frac{arcsin(x)}{x})^{\frac{1}{x^2}} <br /> = \lim_{x \to 0} e^{\frac{ln(\frac{arcsin(x)}{x})}{x^2}}

now let
b =\frac{ln(\frac{arcsin(x)}{x})}{x^2}

if the limit exists, its equal to e^(b), so finding the limit of a is sufficient

thats 0/0 indeterminate, so we can apply L'Hops rule - though i can see it will be a bit messy

 

[lanedance]

though checking on wolfram alpha, a should tend to 1/6
http://www.wolframalpha.com/input/?i=limit+(Arcsin(x)/x)^(1/x^2)

 

[gipc]

sorry, still a no-go. can't get the algebra together. can someone please help? I've applied L'hopital's rule 3 times and it keeps getting uglier.

 

[estro]

I think it might be helpful considering Taylor Polynomials approximation.

 

[gipc]

We didn't learn yet the Taylor thingie. This assignment is about L'Hopital's rule.

 

[lanedance]

ok, so how about starting by looking at the arcsin function and its derivatives, let's abuse the notation a bit and call it a for brevity recognising its a function of x:
a(x) = arcsin(x), \ \ \ \ \ \ \lim_{x \to 0} a(x) = 0
a&#039;(x) = (1-x^2)^{1/2}, \ \ \ \ \lim_{x \to 0} a&#039;(x) = 1
a&#039;&#039;(x) = x(1-x^2)^{3/2}, \ \ \ \lim_{x \to 0} a&#039;&#039;(x) = 0
a&#039;&#039;&#039;(x) = (1-x^2)^{3/2} -3x(1-x^2)^{5/2}, \ \lim_{x \to 0} a&#039;&#039;&#039;(x) = 0

 

[lanedance]

now going back to
b = \lim_{x \to 0}\frac{ln(\frac{arcsin(x)}{x})}{x^2}<br /> = \lim_{x \to 0} \frac{ln(a) - ln(x)}{x^2}<br />

this is 0/0 so using L'Hop
<br /> = \lim_{x \to 0} \frac{a&#039;/a - 1/x}{x^2} = \lim_{x \to 0}\frac{1}{2} \frac{a&#039;x - a}{ax^2}<br />

once again, this is 0/0 so using L'Hop
<br /> = \lim_{x \to 0}\frac{1}{2} \frac{a&#039;&#039;x}{a&#039;x^2+ 2ax}= \lim_{x \to 0}\frac{1}{2} \frac{a&#039;&#039;}{a&#039;x+ 2a}<br />

one more time, this is 0/0 so using L'Hop
<br /> = \lim_{x \to 0}\frac{1}{2} \frac{a&#039;&#039;&#039;}{a&#039;&#039;x+a&#039;+ 2a&#039;}= \lim_{x \to 0}\frac{1}{2} \frac{a&#039;&#039;&#039;}{a&#039;&#039;x+3a&#039;}<br />

and at this point you should be able to sub in with the properties of the derivatives