anonymous
  • anonymous
Find the limit of ln(y^2+2y)/ln(y) as x approaches to 0+.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
As \(x\to0^+\) ?
anonymous
  • anonymous
I got y(2y+2)/(y^2+2y)=2y^2+2y/(y^2+2y)=2/1=2 but the answer is 1. How so?
anonymous
  • anonymous
I don't think you got what I'm asking. There's no \(x\) in the expression. Are you sure about taking the limit as *x* approaches 0?

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More answers

anonymous
  • anonymous
Sorry, it's y.
anonymous
  • anonymous
Okay, so you get to the point where you have (after applying L'Hopital's rule once) \[\lim_{y\to0^+}\frac{y(2y+2)}{y^2+2y}\] Is that right?
anonymous
  • anonymous
So the answer is 2 for sure, right?
anonymous
  • anonymous
No, Wolfram disagrees: http://www.wolframalpha.com/input/?i=Log%5By%5E2%2B2y%5D%2FLog%5By%5D+as+y+approaches+0
anonymous
  • anonymous
What's the answer and how to get there then?
anonymous
  • anonymous
[2(0) + 2]/(0+2) = 1
anonymous
  • anonymous
Euler, how did you get that?
anonymous
  • anonymous
Right, sorry. I completely forgot about the +2's...
Zarkon
  • Zarkon
\[\lim_{y\to0^+}\frac{2y+2}{y+2}=\frac{0+2}{0+2}=\frac{2}{2}=1\]
anonymous
  • anonymous
\[\lim_{y \rightarrow 0+}\frac{ lny+\ln(y+2) }{ lny }=1+\lim_{y \rightarrow 0+} \frac{ \ln(y+2) }{ lny }=1+\frac{ \ln2 }{ - \infty }=1\]

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