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December 14th, 2016, 02:37 AM   #21
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Quote:
 Originally Posted by agentredlum Try breaking up the math tex into several lines so you can more easily identify where the problem is , like this [M ATH] ...... [/M ATH] [M ATH] ..... [/M ATH] [M ATH] ..... [/M ATH]
I did.
Then I simplified it.

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December 14th, 2016, 02:41 AM   #22
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Quote:
 Originally Posted by agentredlum Now that I think more about it, there is an easier approach ... Would you agree that $\displaystyle h = \sqrt{|h|} \sqrt{|h|} \text{sgn}(h)$ ? If you do, then $\displaystyle \frac{\sqrt{|h|}}{h} = \frac{\sqrt{|h|}}{\sqrt{|h|} \sqrt{|h|} \text{sgn}(h)}$ Now just cancel common factor $\displaystyle \sqrt{|h|}$ from numerator and denominator to get the final result in my previous post. I originally set on a path and followed it, not the quickest.
And I started writing in another window on my phone.
Then used copy paste.
Didn't even see you already simplified it.

Last edited by skipjack; December 14th, 2016 at 10:40 AM.

December 14th, 2016, 02:44 AM   #23
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Quote:
 Originally Posted by Luciferis Which could have been realised more easily with. $\displaystyle \frac{\sqrt{|h|}}{h} = \frac{\sqrt{|h|}}{|h|\text{sgn}(h)} = \frac{\sqrt{|h|}}{\sqrt{|h|}\sqrt{|h|}\text{sgn}(h )} = \frac{1}{\sqrt{|h|}\text{sgn}(h)}$
I agree, nice presentation.

Note that the introduction of the sgn(h) notation makes it absolutely clear that the 2-sided limit does not exist since h --> 0 from both the positive and the negative.

Last edited by skipjack; December 14th, 2016 at 10:38 AM.

 Tags abs, calculus, hard, limit, newton, point, problem, quotient, variable

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