Register to reply

Limits- using L'hopital's rule

by Jenkz
Tags: lhopital, limits, rule
Share this thread:
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.
Phys.Org News Partner Science news on
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
Apr4-10, 05:51 PM
P: 1,398
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]
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.

Apr4-10, 06:06 PM
P: 21,311
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.
Apr5-10, 07:06 AM
P: 59
@Mark44: okies, noted.

I've tried differentiating it again and I get:


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 ?


Register to reply

Related Discussions
Do I even need to use L'Hopital's Rule for this Calculus & Beyond Homework 3
L'H˘pital's rule Calculus & Beyond Homework 1
L'Hopital's Rule Calculus & Beyond Homework 6
L'Hopital's Rule Calculus & Beyond Homework 8