# L'Hospital

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• Jul 2nd 2008, 10:27 AM
Mathstud28
Quote:

Originally Posted by Apprentice123
Can you explain why?

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

I think you mean the impication

$\ln(y)=-1$

So exponentiating both sides gives

$e^{\ln(y)}=y=e^{-1}=\frac{1}{e}$
• Jul 2nd 2008, 10:29 AM
Chris L T521
Quote:

Originally Posted by Mathstud28
Are you sure Maclaurin discovered Cramer's rule...I thought it was some monk...

My trusty book tells me that Maclaurin the the original discoverer of Cramer's Rule...

...which monk? (Thinking)
• Jul 2nd 2008, 10:33 AM
Mathstud28
Quote:

Originally Posted by Chris L T521
My trusty book tells me that Maclaurin the the original discoverer of Cramer's Rule...

...which monk? (Thinking)

I don't have that book anymore (Crying)...Cramer? (Smirk)
• Jul 2nd 2008, 10:37 AM
Apprentice123
Quote:

Originally Posted by Mathstud28
Explain the implication? Or explain why you may interchange the limit and the natural log.

I think you mean the impication

$\ln(y)=-1$

So exponentiating both sides gives

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

Clearly. Thank you
• Jul 2nd 2008, 11:18 AM
colby2152
Quote:

Originally Posted by Mathstud28
I don't have that book anymore (Crying)...Cramer? (Smirk)

Cramer's Rule is the method to solve for the discriminant of a non 2x2 matrix.
• Jul 2nd 2008, 11:28 AM
Mathstud28
Quote:

Originally Posted by colby2152
Cramer's Rule is the method to solve for the discriminant of a non 2x2 matrix.

Yeah I know...I was saying "...Cramer?" Meaning that is probably who invented the method.
• Jul 2nd 2008, 02:38 PM
galactus
We could perhaps try it without the Hospital rule.

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

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

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

Probably wasn't worth posting, but I had to post at least one thing today(Hi)
• Jul 2nd 2008, 02:54 PM
Mathstud28
Quote:

Originally Posted by galactus
We could perhaps try it without the Hospital rule.

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

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

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

Probably wasn't worth posting, but I had to post at least one thing today(Hi)

Trust me, I really wanted to not use L'hopital's but...the heading man..the heading. (Angry)
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