Question

Find the limit of ln(y^2+2y)/ln(y) as x approaches to 0+.

Answer 1

As \(x\to0^+\) ?

Answer 2

I got y(2y+2)/(y^2+2y)=2y^2+2y/(y^2+2y)=2/1=2 but the answer is 1. How so?

Answer 3

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?

Answer 4

Sorry, it's y.

Answer 5

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?

Answer 6

So the answer is 2 for sure, right?

Answer 7

No, Wolfram disagrees: http://www.wolframalpha.com/input/?i=Log%5By%5E2%2B2y%5D%2FLog%5By%5D+as+y+approaches+0

Answer 8

What's the answer and how to get there then?

Answer 9

[2(0) + 2]/(0+2) = 1

Answer 10

Euler, how did you get that?

Answer 11

Right, sorry. I completely forgot about the +2's...

Answer 12

\[\lim_{y\to0^+}\frac{2y+2}{y+2}=\frac{0+2}{0+2}=\frac{2}{2}=1\]

Answer 13

\[\lim_{y \rightarrow 0+}\frac{ lny+\ln(y+2) }{ lny }=1+\lim_{y \rightarrow 0+} \frac{ \ln(y+2) }{ lny }=1+\frac{ \ln2 }{ - \infty }=1\]