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July 31st, 2017, 10:56 AM   #1
pst
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Indeterminate form limit

Quick question:

Wouldn't $\displaystyle \lim_{(x,0)\rightarrow(0,0)}\frac{x^2}{x^4} = \frac{0}{0}$?
If so, how did he get the third equation? If we simplify the second equation, we'd have
\frac{1}{x^2} which would be undefined, no?

Last edited by skipjack; July 31st, 2017 at 11:47 AM.
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July 31st, 2017, 11:50 AM   #2
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Was there any relevant context?
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July 31st, 2017, 01:11 PM   #3
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The problem was just a demonstration of how to take limits with multiple variables. By substituting 0 for y, we're testing what the limit would be as we approach from the x-axis so we can see if the limit DNE.

Although I believe I understand now where the author was going with this.

In the second equation,
$\displaystyle lim_{(x,0)\rightarrow(0,0)}\frac{x^2(0)^4}{x^4+3(0 )^4}$
$\displaystyle = lim_{(x,0)\rightarrow(0,0)}\frac{0}{x^4}$
$\displaystyle = 0$

We'd only plug 'n chug the numbers if the function was continuous at 0, but because it isn't, my assumption with the limit being 0/0 arises from false assumption that we would take x=0 and then evaluate
$\displaystyle \frac{(0)^2(0)^2}{(0)^4+3(0)^4} = \frac{0}{0}$

Last edited by pst; July 31st, 2017 at 01:18 PM. Reason: rogue parenthesis
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July 31st, 2017, 01:18 PM   #4
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$\displaystyle \frac{x^2}{x^4}=\frac{1}{x^2}$ which becomes infinite as x->0.
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July 31st, 2017, 01:19 PM   #5
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Quote:
Originally Posted by mathman View Post
$\displaystyle \frac{x^2}{x^4}=\frac{1}{x^2}$ which becomes infinite as x->0.
Right, but the 0 in the numerator would cancel out the x^2 term, right? So we wouldn't be able to take it out of x^4 in the denominator, yes?
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August 1st, 2017, 02:29 PM   #6
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Quote:
Originally Posted by pst View Post
Right, but the 0 in the numerator would cancel out the x^2 term, right? So we wouldn't be able to take it out of x^4 in the denominator, yes?
We are trying to get a limit as x->0,so setting x=0 defeats the purpose.
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