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Limits using L'hopital's rule 
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#1
Apr410, 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. 


#2
Apr410, 05:51 PM

P: 1,403

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] 


#3
Apr410, 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. 


#4
Apr410, 06:06 PM

Mentor
P: 21,397

Limits using L'hopital's rule



#5
Apr510, 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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