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Limits- using L'hopital's rule

by Jenkz
Tags: lhopital, limits, rule
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Jenkz
#1
Apr4-10, 05:17 PM
P: 59
1. The problem statement, all variables and given/known data

Lim tan (x^1/2)/ [x (x+1/2)^1/2 ]
x-> 0

3. The attempt at a solution

I have attempted to differentiate both the denominator and numerator seperately but this just seems to complicate the whole equations and I still get a limit of 0.

I had an idea to square everything, in which case I get a limit of 1. However, I do not think [tan (x^1/2)]^2 = tanx

Please help? My friend and I have been trying to work this out. It isn't homework, merely revision and further understanding.
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willem2
#2
Apr4-10, 05:51 PM
P: 1,396
You must have made an error while differentiating, the limit is not 0.

[tan (x^1/2)]^2 is not the same as tanx. Try [itex] x = \pi [/itex]
Jenkz
#3
Apr4-10, 06:03 PM
P: 59
My mistake, the equation is

tan (x^1/2)/ [x (1+1/x)^1/2 ] sorry

I'll try your hint, and have another go at differentiating it. Thanks.

Mark44
#4
Apr4-10, 06:06 PM
Mentor
P: 21,283
Limits- using L'hopital's rule

Quote Quote by Jenkz View Post
My mistake, the equation is
tan (x^1/2)/ [x (x+1/x)^1/2 ]
Minor point - that's not an equation. They're easy to spot because there's one of these- = - in an equation.
Jenkz
#5
Apr5-10, 07:06 AM
P: 59
@Mark44: okies, noted.

I've tried differentiating it again and I get:

[tex]\frac{\frac{sec^{2}\sqrt{x}}{2\sqrt{x}}}{\sqrt{\frac{1}{x}+1}-\frac{1}{2\sqrt{\frac{1}{x}+x}}}[/tex]

But it still doesn't give me a limit.

I'm not too sure how to use your hint. As if i let [tex]\pi=x[/tex] Doesnt it just mean [tex]\pi[/tex] tends towards 0 instead of x ?

Confused...


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