L'Hopital's with two variables

**blackhug** · August 12th 2010, 03:45 PM

Hi I'm having trouble solving the following limit:

$\lim_{t\to 0}\frac{t}{\sqrt(x + t) - \sqrt(x)}$

I've got as far as applying L'Hopital's rule to the top and bottom of the quotient, getting:
$<br /> \frac{\frac{dt}{dt}}{\frac{1 + \frac{dt}{dt}}{2\sqrt(x + t)}-\frac{1}{2\sqrt(x)}}$

As far as I know, this is the correct step to take. I'm clueless as to what I should do from here.

**roninpro** · August 12th 2010, 04:03 PM

This is simply the reciprocal of the difference quotient

$f'(x) = \displaystyle \lim_{t\to 0} \frac{f(x+t)-f(x)}{t}$

where $\displaystyle f(x)=\sqrt{x}$ .

Therefore, $\displaystyle \lim_{t\to 0} \frac{t}{\sqrt{x+t}-\sqrt{x}}=\frac{1}{f'(x)}=2\sqrt{x}$ .

**blackhug** · August 12th 2010, 04:46 PM

Thanks for the response. Can it be done using L'Hopital's rule though? Because the question specifically states that I need to use the rule.

**chiph588@** · August 12th 2010, 08:05 PM

Originally Posted by blackhug

Thanks for the response. Can it be done using L'Hopital's rule though? Because the question specifically states that I need to use the rule.

Yes it can. You have the right idea but you took the derivative of the denominator wrong.

$\displaystyle \frac d{dt}\left(\sqrt{x+t}-\sqrt{x}\right) = \frac1{2\sqrt{x+t}}$ since we're treating $x$ as a constant.

**Vlasev** · August 12th 2010, 08:16 PM

Also

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taking the limit is now trivial!

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