Question

lim x->16 (4-sqrt(x))/(s-16)

Answer 1

is it supposed to be s on bottom?

Answer 2

yeah

Answer 3

\[\frac{ 4-\sqrt{s} }{ s-16 }\]

Answer 4

Oh, so its supposed tobe as "s" goes to 16, not x.

Answer 5

oh yeah, sorry

Answer 6

No guarantees itll work, but when you see a square root +/- something, its best to try multiplying top and bottomby the conjugate first.

Answer 7

Yeah, i did that. I just don't know how to factor out (16-s)/(s-16)(4+sqrt(s))

Answer 8

for (16-s)/[(s-16)(4+sqrt(s))] simply distribute the denominator as you would any algebraic expression, and then see what further simplification you can do.

Answer 9

an easier way would be to factor out the bottom as well

Answer 10

Hmm...maybe not/

Answer 11

\[\frac{ -(s-16) }{ (s-16)(4+\sqrt{s}) }\]

Answer 12

that'd still be dividing by 0

Answer 13

Because (16-s) = (-s+16) = -(s-16)

Answer 14

so factor out the negative from top then, not denominator

Answer 15

Doesnt matter honestly.

Answer 16

\(\huge \frac{(4-\sqrt s)}{(s-16)}=\frac{4-\sqrt s}{(\sqrt s-4)(\sqrt s+4)}=-\frac{\sqrt s-4}{(\sqrt s-4)(\sqrt s+4)}\)

Answer 17

\[\frac{ (16-s) }{-(16-s)(4+\sqrt{s}) }\]

Answer 18

i'm getting -1/8

Answer 19

Yep :3

Answer 20

And what inky did is fine, too, he just did a difference of squares to factor it.

Answer 21

so pretty much with limits; anything you can think of to make it not 0/0 or x/0

Answer 22

Basically. There are a lot of little tricks you can do, but if I ever see sqrt +/- something then conjugate is first thought.

Answer 23

The problem of having to rationalize the denominator is classic for limits. It pops up almost any class to give the problem of needing to multiply by the conjugate - little algebra fun that everyone loves

Answer 24

Sometimes you may have to rationalize the denominator, sometimes the numerator, sometimes even both at the same time.

Answer 25

Yeah, I just didn't think to factor out the negative.

Answer 26

you will learn, in the future, ways to resolve indeterminate expressions for limits - for now though, stick to simplification and trying get rid of determinant forms

Answer 27

Yeah, calc 2 gives ya a trick to deal with anything that ends up as 0/0 or infinity/infinity.

Answer 28

I have another one if you guys are interested.

Answer 29

Go for it.

Answer 30

Post as new question :)

Answer 31

lim x ->9 sqrt(x)/(x-9)^4

Answer 32

oh okay i will