# Limits- using L'hopital's rule

## Homework Statement

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

## 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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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 $x = \pi$

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
Mentor
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.

@Mark44: okies, noted.

I've tried differentiating it again and I get:

$$\frac{\frac{sec^{2}\sqrt{x}}{2\sqrt{x}}}{\sqrt{\frac{1}{x}+1}-\frac{1}{2\sqrt{\frac{1}{x}+x}}}$$

But it still doesn't give me a limit.

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

Confused...