1. ## dy/dx question...

Hey all,

I'm getting ready for my first semester of Calculus and I'm a big dork so I'm actually interested in the subject.

Anyway I have a quick question: when differentiating $\sqrt{x}$ I was taught by my trusty textbook to multiply by the conjugate using the difference quotient and so then after several calculations you end up with:

$\lim_{x \rightarrow 0}\frac{1}{\sqrt{x + h} + \sqrt{x}}$

THEN for some reason you can get rid of the "h" in $\sqrt{x + h}$ and you're left with $\frac{1}{2\sqrt{x}}$.

Now I get how to do everything just fine, I just want to know why the "h" can just go away like that.

Thanks for the help!

2. There is a mistake in your limit:
$\lim_{x \rightarrow 0}\frac{1}{\sqrt{x + h} + \sqrt{x}}$

should be:
$\lim_{h \rightarrow 0}\frac{1}{\sqrt{x + h} + \sqrt{x}}$

Do you now see why the result follows?

3. Ah yes I see I messed up the limit and should have put as "h" tends to 0 but I don't know enough about limits yet to understand why the "h" can go away.

4. Hello,
Originally Posted by DaveDammit
Ah yes I see I messed up the limit and should have put as "h" tends to 0 but I don't know enough about limits yet to understand why the "h" can go away.
Consider this (only if there is no undetermined form, that is infinity x 0, 0/0, infinity/infinity, etc... you'll see it later)

h goes to 0
so x+h goes to x+0=x
so sqrt(x+h) goes to sqrt(x) !

that's the basic idea

$\lim_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}}=\frac{1}{\lim\limits _{h \to 0} \sqrt{x+h}+\sqrt{x}}=\dots$

5. Gotcha. Thanks to the both of you! Very helpful.