Question

help me step by step?

lim as x approaches 8 of (sqrt(x+1)-3)/(x-8)

Answer 1

I know I can't factor, so what do I do now? Rationalize?

Answer 2

\[\large \lim_{x \rightarrow 8} \frac{\sqrt{x+1}-3}{x-8}\]Hmmmm so right now we're getting an indeterminate form...
So we need to find a way to cancel something out.
Yesss! The way we'll do that, is by multiplying the top and bottom by the CONJUGATE of the top. Do you know what that will look like? :D

Answer 3

\[\frac{ \sqrt{x+1} -3}{ x-8}*\frac{ \sqrt{x+1} +3}{ \sqrt{x+1} +3} ?\]

Answer 4

Yes, good good. Let's simply the top, but DON'T multiply out the bottom, just leave the 2 terms separate for now.

Answer 5

doesn't the top cancel out?

Answer 6

\[\frac{ 1 }{ \sqrt{x+1}+3*x-8}\] ?

Answer 7

Multiplying conjugates will give you the - Difference of Squares. I'm not sure why you think they would cancel out :D heh

Answer 8

err.. sorry ;/

Answer 9

\[\large (a-b)(a+b)=a^2-b^2\]

Answer 10

\[\sqrt{x+1}^2-3^2?\]

Answer 11

Yes c:

Answer 12

\[\frac{ \sqrt{x+1}^2-3^2 }{ \sqrt{x+1}+3*x-8 }\]

Answer 13

Use parentheses on the bottom :) It looks too sloppy that way.
And simplify the sqrt term, see how you have a sqrt root being squared?

Answer 14

\[\frac{ x-8 }{ (\sqrt{x+1}+3)(x-8) }= \sqrt{x+1}+3\]

Answer 15

Yay looks good! Now it's in a nice form where we can take the limit!
Since we were able to cancel some garbage out.

Answer 16

=6?

Answer 17

Sounds about right :D yay team.

Answer 18

it says 1/6 as the answer? did I mess up somewhere

Answer 19

oh lol sorry i wasn't paying attention :)

Answer 20

When you made that nice cancellation. How did the thing in the denominator pop up to the top?? :O

Answer 21

when I cancel it needs to become 1/.... ?

Answer 22

Yah :3

Answer 23

x-8 / x-8 = 1

Answer 24

gotcha ;D thanks!!