Question

Square root of (x+deltaX)+ 2 - square root of (x-2)/ delta x

Answer 1

What is the question?

Answer 2

Find the lim as delta x approaches 0

Answer 3

Do you mean
\[\sqrt{x+\Delta x}+2-\frac{\sqrt{x-2}}{\Delta x}\]

Answer 4

\[\frac{\sqrt{x+\Delta x}+2-\sqrt{x+2}}{\Delta x}\]

Answer 5

yes

Answer 6

but for the first sqrt its (x+deltax)+2

Answer 7

Ok this is an indetermination of the type 0/0. You can use this method,
\[\frac{\sqrt{x+\Delta x+2}-\sqrt{x+2}}{\Delta x}\cdot\frac{\sqrt{x+\Delta x+2}+\sqrt{x+2}}{\sqrt{x+\Delta x+2}+\sqrt{x+2}}\]

Answer 8

the answer is apparently \[\frac{ 1 }{ 2\sqrt{x+2}}\]

Answer 9

alright ill try it

Answer 10

Once you multiply the factors, you'll obtain,
\[\frac{\Delta x}{\Delta x(\sqrt{x+\Delta x+2}+\sqrt{x+2})}=\frac{1}{(\sqrt{x+\Delta x+2}+\sqrt{x+2})}\]And yes, now take the limit Delta x tends to 0 and you'll obtain the answer.

Answer 11

;)

Answer 12

yup it worked perfectly

Answer 13

Use this "trick" when you have limits of the same type (also if you see derivatives, you'll see that it is possible use the L'Hopital rule)

Answer 14

we havent gotten to derivatives in ap calc yet so yeah... ahah