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January 10th, 2017, 08:05 AM  #1 
Member Joined: Jan 2017 From: California Posts: 80 Thanks: 8  radius of convergence with indeterminate form 20170110_080156.jpg The limit of this ratio test for this series suggests that for any x except zero the series diverge. The thing I cannot get over is the following: What if I choose an x that is so infinitely close to zero such as it dominates infinity? Such as that the product of the two factors is less than 1. I believe this was the case of an indeterminate form when L'Hôpital's rule was necessary in limits calculations. Thanks in advance. Last edited by skipjack; January 10th, 2017 at 08:46 AM. 
January 10th, 2017, 08:57 AM  #2  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,872 Thanks: 766  Quote:
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January 10th, 2017, 09:02 AM  #3 
Member Joined: Jan 2017 From: California Posts: 80 Thanks: 8 
What if I choose an x so small such as 0.00000.... with the 1 all the way to the north pole? Can I do that? I assume I could since the value of x could be continuous and not discrete. Couldn't that small number overpower infinity (0 vs infinity) type of fight?
Last edited by skipjack; January 10th, 2017 at 10:50 AM. 
January 10th, 2017, 10:01 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,084 Thanks: 2360 Math Focus: Mainly analysis and algebra 
Two approaches to put you on the right track

January 10th, 2017, 10:25 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,872 Thanks: 766  Can you quantify the number of "0"s required to go "all the way to the north pole"? You can say a very large, but specific, number of "0"s. However, that would not "overpower" infinity. Any number "times infinity" is "infinity". I put "times infinity" and "infinity" in quotes because "infinity" is NOT a real number and multiplication of any real number with "infinity" is not defined.
Last edited by skipjack; January 10th, 2017 at 10:52 AM. 
January 10th, 2017, 11:21 AM  #6  
Member Joined: Jan 2017 From: California Posts: 80 Thanks: 8  Quote:
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However, there was an indeterminate case that forced the use of L'Hôpital's in limits calculation. It was stating that 0+ multiplied by infinity is not equal to infinity and that's why we had to rearrange things and present them in the form 0/0 or infinity over infinity. I feel like this is contradicting the stuff we're discussing. I appreciate your help guys. Last edited by skipjack; January 10th, 2017 at 11:29 AM.  
January 10th, 2017, 12:06 PM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,084 Thanks: 2360 Math Focus: Mainly analysis and algebra 
If $f(x) \to 0$ and $g(x) \to \infty$ then $$\lim_{x \to a} f(x)g(x)$$ is the indeterminate case $0 \cdot \infty$. This is the case you are thinking of. But that is because the limit of $f(x)$ is equal to zero, not because $f$ is very small. But if $f(x)=c \ne 0$ and $g(x) \to \infty$ then $$\lim_{x \to a} f(x)g(x)$$ is the undefined (but not indeterminate) case $c \cdot \infty = \pm\infty$. And if $f(x)=0$ and $g(x) \to \infty$ then $$\lim_{x \to a} f(x)g(x) = \lim_{x \to a} 0 \cdot g(x) = \lim_{x \to a} 0 = 0$$ 
January 10th, 2017, 12:28 PM  #8  
Member Joined: Jan 2017 From: California Posts: 80 Thanks: 8  Quote:
 

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convergence, form, indeterminate, radius 
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