Explain the implication? Or explain why you may interchange the limit and the natural log.

I think you mean the impication

$\displaystyle \ln(y)=-1$

So exponentiating both sides gives

$\displaystyle e^{\ln(y)}=y=e^{-1}=\frac{1}{e}$

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- Jul 2nd 2008, 10:27 AMMathstud28
- Jul 2nd 2008, 10:29 AMChris L T521
- Jul 2nd 2008, 10:33 AMMathstud28
- Jul 2nd 2008, 10:37 AMApprentice123
- Jul 2nd 2008, 11:18 AMcolby2152
Cramer's Rule is the method to solve for the discriminant of a non 2x2 matrix.

- Jul 2nd 2008, 11:28 AMMathstud28
- Jul 2nd 2008, 02:38 PMgalactus
We could perhaps try it without the Hospital rule.

Let $\displaystyle t=\frac{1}{1-x}, \;\ x=1-\frac{1}{t}$

Then we get the recognizable and fairly famous limit which can be solved by a suitable substitution:

$\displaystyle \lim_{t\to {\infty}}\left(1-\frac{1}{t}\right)^{t}=\frac{1}{e}$

Probably wasn't worth posting, but I had to post at least one thing today(Hi) - Jul 2nd 2008, 02:54 PMMathstud28