$\displaystyle

=\lim _{x->0} \frac{\frac{1}{\sqrt{a+x+2}} - \frac{1}{\sqrt{a+x}}}{x}

$

could you please help me to solve this problem please. Thank you.

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- Feb 28th 2009, 09:54 PMthangbelimit problem
$\displaystyle

=\lim _{x->0} \frac{\frac{1}{\sqrt{a+x+2}} - \frac{1}{\sqrt{a+x}}}{x}

$

could you please help me to solve this problem please. Thank you.

- Mar 1st 2009, 03:36 AMmr fantastic
Several approaches are possible.

There's a simple one if you know differential calculus ..... you can consider the function $\displaystyle f(a) = \frac{1}{\sqrt{a + 2}}$. Then, with a (perhaps disquieting) change from the usual notation ($\displaystyle x$ instead of $\displaystyle h$, $\displaystyle a$ instead of $\displaystyle x$) you can consider the derivative of $\displaystyle f(a)$ from first principles:

$\displaystyle f'(a) = \lim_{x \rightarrow 0} \frac{f(a + x) - f(a)}{x}$.

So the answer will be the derivative of f(a) with respect to a .... - Mar 1st 2009, 04:26 AMHallsofIvy
- Mar 1st 2009, 04:30 AMmr fantastic
- Mar 1st 2009, 05:06 PMthangbe
the limit is exists and the answer is.

$\displaystyle

=\lim _{x->0} \frac{\frac{1}{\sqrt{a+x+2}} - \frac{1}{\sqrt{a+x}}}{x} =\frac {-1}{2{(a+2)}^\frac{3}{2}}

$

but I don't know how to show for it.

- Mar 1st 2009, 05:31 PMmr fantastic
- Mar 1st 2009, 06:33 PMthangbe
I understand in your post #2 that why I get that answer. However my teacher didn't allows us to use that menthod :( thank you for you help sir.

- Mar 1st 2009, 06:41 PMmatheagle
Clearly there is a typo.

But once that is resolved, my favourite way to solve these type of limits is by multiplying by the conjugate. Not many people teach that these days. - Mar 1st 2009, 06:56 PMmr fantastic
- Mar 1st 2009, 07:05 PMmatheagle
I agree. I'm helping someone right now.

But substituting the math and typing tex at the same is tough for me.

I'm afraid I'll do a stupid calculation and I cannot edit after 15 minutes.

Leaving forever with a dumb math error.

That would be embarassing.

I really like the conjugate technique. - Mar 1st 2009, 07:18 PMthangbe
uhm... could you just simply tell me the procedure.? or just give a simple example. So that I can figure it out.... I was stuck in how to get x out of the squareroot value.

- Mar 2nd 2009, 01:49 AMmr fantastic
Note that $\displaystyle \frac{1}{\sqrt{a + x + 2}} - \frac{1}{\sqrt{a+2}} = \frac{\sqrt{a+2} - \sqrt{a+x+2}}{\sqrt{a+x+2} \, \sqrt{a+2}}$.

Therefore:

$\displaystyle \frac{\frac{1}{\sqrt{a + x + 2}} - \frac{1}{\sqrt{a+2}}}{x} = \frac{\sqrt{a+2} - \sqrt{a+x+2}}{x \sqrt{a+x+2} \, \sqrt{a+2}}$

$\displaystyle = \frac{(\sqrt{a+2} - \sqrt{a+x+2}) \cdot (\sqrt{a+2} + \sqrt{a+x+2})}{x \sqrt{a+x+2} \, \sqrt{a+2} \cdot (\sqrt{a+2} + \sqrt{a+x+2})}$

$\displaystyle = \frac{(a + 2) - (a + x + 2)}{x \sqrt{a+x+2} \, \sqrt{a+2} \cdot (\sqrt{a+2} + \sqrt{a+x+2})}$

$\displaystyle = \frac{-x}{x \sqrt{a+x+2} \, \sqrt{a+2} \cdot (\sqrt{a+2} + \sqrt{a+x+2})}$

$\displaystyle = \frac{-1}{\sqrt{a+x+2} \, \sqrt{a+2} \cdot (\sqrt{a+2} + \sqrt{a+x+2})}$.

Now take the limit $\displaystyle x \rightarrow 0$. - Mar 2nd 2009, 06:07 AMthangbe
thank you so much for helping :)