# Thread: L'Hopital's with two variables

1. ## L'Hopital's with two variables

Hi I'm having trouble solving the following limit:

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

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

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

2. This is simply the reciprocal of the difference quotient

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

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

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

3. 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.

4. 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.

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

5. Also

$\displaystyle \displaystyle \frac{t}{\sqrt(x + t) - \sqrt(x)} = \frac{t(\sqrt{x+t}+\sqrt{x})}{(\sqrt(x + t) - \sqrt(x))(\sqrt(x + t) + \sqrt(x))} = \frac{t(\sqrt{x+t}+\sqrt{x})}{x+t-x} = \sqrt{x+t}+\sqrt{x}$

taking the limit is now trivial!