Question

Evaluate the limit as x approaches 0 for

{3-(9-x)^.5}/x

Answer 1

Factor an x from the top and bottom and cancel them out.

Answer 2

use LHopital's rule

Answer 3

Oh, sorry I misread. Can't get that x out of the parens

Answer 4

differentiate numerator and denominator separately

Answer 5

that x on the top is inside of a radical

Answer 6

1/(2sqrt(9-x) will be the result. then plug x = 0
1/(2sqrt(9) = 1/6

Answer 7

HOLD ON Kdabr what calculus are you in.... helping tutors, he can't use l'hopitals unless he's in calc 2

Answer 8

or unles he's been taught it

Answer 9

Im in precal haha, about to enter calc BC

Answer 10

So probably not l'hopitals since he's just getting into limits

Answer 11

helping tutors got the right answer but I have no idea what he did

Answer 12

here kdabr... you're goingto have to rationalize it.. and yes kdabr, he used l'hopitals rule which is taught after calculus 1 as there is more rules with it that can make it tricky..I have a problem much ilke it in my book i'll send you it now

Answer 13

I have use the L'hopital's rule.. and since you are in precalc... rationalisation could work better for your grade

Answer 14

that page looks spot on

Answer 15

your problem is very similar to this problem if you were to mimic what it looks like in my book you don't necessarily have to tho

Answer 16

I tried doing that already and ended up with x/(3x+sqrt(9-x))

Answer 17

now if you want to dabble into l'hopitals to perhaps check yourself i could send you that also but if you were to solve usinng L'H you'd probably get 0 pts

Answer 18

you are correct about that outkast

Answer 19

and also it does not work in all cases only certain indeterminate forms

Answer 20

i'm hoping polpak is doing the problem

Answer 21

as i amnot lol

Answer 22

\[\lim_{x \rightarrow 0}\frac{3-\sqrt{9-x}}{x} \]\[= \lim_{x\rightarrow 0}\frac{3-\sqrt{9-x}}{x} \cdot \frac{3+\sqrt{9-x}}{3+\sqrt{9-x}}\]\[=\lim_{x\rightarrow0}\frac{9-(9-x)}{x(3+\sqrt{9-x})}\]\[=\lim_{x\rightarrow0} \frac{9-9 + x}{x(3+\sqrt{9-x})}\]\[=\lim_{x\rightarrow0} \frac{x}{x(3+\sqrt{9-x})}\]\[=\lim_{x\rightarrow0} \frac{1}{3+\sqrt{9-x}}\]\[= \frac{1}{3 + \sqrt{9 + 0}} = \frac{1}{3+3} = \frac{1}{6}\]

Answer 23

yes indeed what he did you didn't pull out an x and cancel the top... it should all cancel

Answer 24

You forgot your factor of x on the bottom

Answer 25

but why do those x's become ones?

Answer 26

to cancel with the x on the top.

Answer 27

oh of course

Answer 28

Because x/x = 1 as long as x isn't 0 and we're approaching 0, we don't actually care what happens when we get there.

Answer 29

did you want l'hopitals for a check or idk even if i'td make sense to you as you probably dont know how to differentiate?

Answer 30

I think he has the answer in a key. Just didn't know how to find it.