anonymous
  • anonymous
please please please help with a limits question!!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
i'm studying limits now in calc bc (just started) maybe i can help you
anonymous
  • anonymous
\[\lim_{x \rightarrow 0} (\sqrt{x+9}-3)\div(x)\]
anonymous
  • anonymous
the limit of both of those, or just the thing in the first parenth and hen that divided by the 2nd parenth?

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anonymous
  • anonymous
i know it ends up being 1/6 but im not sure why
anonymous
  • anonymous
the two parentheses are the equation and its the limit as x approaches that
anonymous
  • anonymous
so you mean \[\lim_{x \rightarrow 0} \frac{ \sqrt{x+9} }{ x }\]
anonymous
  • anonymous
sorry -3 in the numerator too
anonymous
  • anonymous
yeah lol im not very good at using that
anonymous
  • anonymous
i feel like the x in the denominator would make the whole thing undefined... hmmm...
anonymous
  • anonymous
\[\lim_{x \rightarrow 0} \frac{ \sqrt{x+9}-3 }{ x }\]
anonymous
  • anonymous
yay i did it
anonymous
  • anonymous
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.
anonymous
  • anonymous
well thats why i multiplied the top and the bottom by the conjugate of the top
anonymous
  • anonymous
no idea. didnt learn enough calc to know that.
anonymous
  • anonymous
so take the lim of 1/2(x+9)^-1/2?? @asdafogbucket
anonymous
  • anonymous
Yeah, that ends up giving you 9^(-1/ 2) / 2 since you can substitute now that the x in the denominator is gone.
anonymous
  • anonymous
ohhh ok well thats cool mrs. carder didn't teach us that rule
anonymous
  • anonymous
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.
anonymous
  • anonymous
well thanks youre a really big help im in AP calc AB
anonymous
  • anonymous
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?
anonymous
  • anonymous
We literally have had class for 1 day.
anonymous
  • anonymous
@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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
by multiplying by the conjugate but it didnt work for me so idk @Avihirschx
anonymous
  • anonymous
@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 \[\frac{ 1 }{ \sqrt{9} + 3 } = \frac{ 1 }{ 3 + 3 } = \frac{ 1 }{6 }\]

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