Question

please please please help with a limits question!!!

Answer 1

i'm studying limits now in calc bc (just started) maybe i can help you

Answer 2

\[\lim_{x \rightarrow 0} (\sqrt{x+9}-3)\div(x)\]

Answer 3

the limit of both of those, or just the thing in the first parenth and hen that divided by the 2nd parenth?

Answer 4

i know it ends up being 1/6 but im not sure why

Answer 5

the two parentheses are the equation and its the limit as x approaches that

Answer 6

so you mean \[\lim_{x \rightarrow 0} \frac{ \sqrt{x+9} }{ x }\]

Answer 7

sorry -3 in the numerator too

Answer 8

yeah lol im not very good at using that

Answer 9

i feel like the x in the denominator would make the whole thing undefined... hmmm...

Answer 10

\[\lim_{x \rightarrow 0} \frac{ \sqrt{x+9}-3 }{ x }\]

Answer 11

yay i did it

Answer 12

An easy way to do it is to use L'Hopital's rule, if you know it. Because you get division by zero when setting x equal to 0, you can take the derivative of the top and bottom and then take the limit of that.

Answer 13

well thats why i multiplied the top and the bottom by the conjugate of the top

Answer 14

no idea. didnt learn enough calc to know that.

Answer 15

so take the lim of 1/2(x+9)^-1/2?? @asdafogbucket

Answer 16

Yeah, that ends up giving you 9^(-1/ 2) / 2 since you can substitute now that the x in the denominator is gone.

Answer 17

ohhh ok well thats cool mrs. carder didn't teach us that rule

Answer 18

It doesn't come up until Calculus BC sometimes despite how simple and useful it is. You can do that for any limit that causes division by zero.

Answer 19

well thanks youre a really big help im in AP calc AB

Answer 20

So i'm in calc bc but we just started so i dont know derivatives yet... any way to do this w/o using derivatives?

Answer 21

We literally have had class for 1 day.

Answer 22

@Avihirschx If you're in BC I would assume you've taken AB which would mean you know derivatives... But you should be able to do it algebraically like she mentioned earlier too.

Answer 23

In my school in senior year there's an option to take ab calc or ab and bc in one year (it's rushed) which is called bc. and how did she do it algebraically? i didnt catch that.

Answer 24

by multiplying by the conjugate but it didnt work for me so idk @Avihirschx

Answer 25

@Avihirschx Multiplying by the conjugate does work, it's just a lot more writing.
\[\frac{ \sqrt{x + 9} - 3 }{ x } * \frac{ \sqrt{x + 9} + 3 }{ \sqrt{x + 9} + 3 } = \frac{ (\sqrt{x + 9} - 3)(\sqrt{x + 9} + 3) }{ x(\sqrt{x + 9} + 3) }\]
\[\frac{ (\sqrt{x + 9} - 3)(\sqrt{x + 9} + 3) }{ x(\sqrt{x + 9} + 3) } = \frac{(x + 9) - 9 }{ x(\sqrt{x + 9} + 3) } = \frac{x }{ x(\sqrt{x + 9} + 3) }\]
\[\frac{ x }{ x(\sqrt{x + 9} + 3) } = \frac{ 1 }{ \sqrt{x + 9} + 3 }\]
Substituting 0 for x then gives
\[\frac{ 1 }{ \sqrt{9} + 3 } = \frac{ 1 }{ 3 + 3 } = \frac{ 1 }{6 }\]