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• Thread starter Jason76
• Start date Mar 1, 2013

J

Jason76

Senior Member

Joined Oct 19, 2012 Messages 1,180 The ln of 0 is infinity. Take this example: The Limit as x approaches 0 from the right (positive side) of \(\displaystyle \dfrac{lnx}{x^{-1}}\) So the top would be infinity as 0 is plugged in, but the bottom would be 0. So what is infinity divided by 0? But the book says the bottom is positive infinity, while the top is negative infinity, not just infinity. Why? Once you actually use L'Hopital's rule then by taking the derivative you get a limit of 0. Last edited: Mar 1, 2013 H

HallsofIvy

Elite Member

Joined Jan 27, 2012 Messages 7,760

Jason76 said: The ln of 0 is infinity. Take this example: Click to expand...

No, the logarithm of 0 (to any base) does not exist. In terms of the limit we might say that ln(x) goes to negative infinity as x goes to 0.

The Limit as x approaches 0 from the right (positive side) of \(\displaystyle \dfrac{lnx}{x^{-1}}\) So the top would be infinity as 0 is plugged in Click to expand...

No, it is not.

, but the bottom would be 0. So what is infinity divided by 0? Click to expand...

It's not anything. You cannot do arithmetic with "infinity" as if it were an actual number.

But the book says the bottom is positive infinity, while the top is negative infinity, not just infinity. Why? Click to expand...

What is ln(0.1)? What is ln(0.001)?

Once you actually use L'Hopital's rule then by taking the derivative you get a limit of 0. Click to expand...

L'Hopital's rule does not apply here. L'Hopital's rule only applies to "0/0" or "infinity/infinity" Last edited: Mar 1, 2013 L

lookagain

Elite Member

Joined Aug 22, 2010 Messages 3,250

HallsofIvy said: L'Hopital's rule does not apply here. L'Hopital's rule only applies to "0/0" or "infinity/infinity" Click to expand...

Jason76, other indeterminate forms that sometimes can have L'Hopital's rule applied, with some algebraic rewriting done first, are: \(\displaystyle 1^{\infty}, \ \ \infty^0, \ \ \infty - \infty, \ \ (0)(\infty), \ \ and \ \ \ 0^0.\) H

HallsofIvy

Elite Member

Joined Jan 27, 2012 Messages 7,760

lookagain said: Jason76, other indeterminate forms that sometimes can have L'Hopital's rule applied, with some algebraic rewriting done first, are: \(\displaystyle 1^{\infty}, \ \ \infty^0, \ \ \infty - \infty, \ \ (0)(\infty), \ \ and \ \ \ 0^0.\) Click to expand...

True and I thought about mentioning that. But L'Hopital's is applied, as you say "with some rewriting done first" to put them into the form "0/0" or "infinity/infinity". Also, Jason76's example was clearly none of those forms. He seemed to think he was getting "infinity/infinity" but \(\displaystyle \lim_{x\to 0} x^{-1}\) is NOT "infinity". You must log in or register to reply here. Share: Facebook X (Twitter) Reddit Pinterest Tumblr WhatsApp Email Share Link

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