Need some help Finding a limit (if it exists even)

**jimmyp** · May 20th 2009, 07:17 PM

Howdy,

I need some help finding this limit:

$lim \Delta x -> 0$

$\sqrt{x+\Delta x}-\sqrt{x}$
______________________________

$\Delta x$

(P.S. Sorry I was having a bit of trouble using the \frac command, just so you know it's suppose to be delta X on the bottom of the fraction as I tried to show using the ___). Thanks.

jimmyp

**TheEmptySet** · May 20th 2009, 08:07 PM

Originally Posted by jimmyp

Howdy,

I need some help finding this limit:

$lim \Delta x -> 0$

$\sqrt{x+\Delta x}-\sqrt{x}$
______________________________

$\Delta x$

(P.S. Sorry I was having a bit of trouble using the \frac command, just so you know it's suppose to be delta X on the bottom of the fraction as I tried to show using the ___). Thanks.

jimmyp

first we need to multiply the numerator and denominator by the conjugate of the numerator

$\lim_{\Delta x \to 0}\frac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}\left( \frac{\sqrt{x+\Delta x}+\sqrt{x}}{\sqrt{x+\Delta x}+\sqrt{x}}\right)$

Simplifying we get

$\lim_{\Delta x \to 0}\frac{\Delta x}{\Delta x(\sqrt{x+\Delta x}+\sqrt{x})}=\lim_{\Delta x \to 0}\frac{1}{(\sqrt{x+\Delta x}+\sqrt{x})}$

Finally taking the limit we get

$\frac{1}{2\sqrt{x}}$

**stapel** · May 22nd 2009, 08:06 AM

Originally Posted by TheEmptySet

first we need to multiply the numerator and denominator by the conjugate of the numerator....

That "multiplying by the conjugate" thing, by the way, is a "trick" you should remember. You will almost certainly need it on the next test!

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