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Can anyone explain this wolframalpha result?

$ f(x)=\lim\limits_{h \to 0} \frac{-1}{(3x-2)^2} = \frac{-1}{(2-3x)^2}$

[lim ((-1)/((3x-2)^2)) as h->0] = [((-1)/((2-3x)^2)) ]

While this is not equal:

$ f(x)=\lim\limits_{h \to 0} \frac{-1}{(3x-2)^2} = \frac{-1}{(3x-2)^2}$

[lim ((-1)/((3x-2)^2)) as h->0] = [((-1)/((3x-2)^2))]

edit: addint $f(x)=$ to stop confusion about the lack of x in the limit.

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Is the $h \to \infty$ intentional? Or is the question about why $(3x-2)^2=(2-3x)^2$ –  Bonnaduck Nov 19 '13 at 0:34
    
I'm pretty sure I wrote ${h \to 0}$. The question is mostly about why the second wolfram link is not considered equal by wolfram. The first one say "Result: True", the second does not. When i remove the limit from both, they are true. –  Dorus Nov 19 '13 at 14:08
1  
Apologies, I meant $h \to 0$. I doubt you'll find a reason as to why WA is not giving a "true-false" response for the second one, however. Note that it is not saying that the second one is unequal. –  Bonnaduck Nov 19 '13 at 17:17
    
Thanks, that pretty much answers my question. –  Dorus Nov 20 '13 at 0:00

2 Answers 2

up vote 1 down vote accepted

You are taking the limit as $h \to 0$, but $h$ does not exist in the limit. Perhaps you meant $ \lim_{x \to 0} \frac{-1}{(3x-2)^2}$, where we are tending $x$ to $0$?

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I did start with h, but i already lost it along the road. If you really want to know, this was my starting point: [lim ( ((2*(x+h)-1)/(3*(x+h) -2) - (2x-1)/(3x-2) ) /h) as h->0] –  Dorus Nov 19 '13 at 14:13
    
Accepting this because you answered my question in the comments up there. –  Dorus Nov 21 '13 at 22:40

If you are asking why $\dfrac{-1}{(3x-2)^2}=\dfrac{-1}{(2-3x)^2}$, this is because $(3x-2)^2=(2-3x)^2$, which is because you can factor out a negative one, twice so they cancel.

$(2-3x)^2=(2-3x)(2-3x)=-(3x-2)(2-3x)=-(-(3x-2)(3x-2))\\=(3x-2)(3x-2)=(3x-2)^2$.


In the case you are asking why Wolfram Alpha returned an incorrect evaluation of the limit, that is most likely because you used $h$ as your variable in the limit instead of $x$.

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Actually the my left hand side was (3x+3h-2)(3x-2), since h lim 0, I removed the h. Can you explain why the second Wolfram link is not true? –  Dorus Nov 19 '13 at 10:29
    
I don't think I understand. What do you mean by the second link is not true? When I click the second link it tells me that $\displaystyle\lim_{h\to 0}-\dfrac{1}{(3x-2)^2}=-\dfrac{1}{(3x-2)^2}$, which is true. Can you please clarify? –  Sujaan Kunalan Nov 19 '13 at 17:59
    
If you are referring to the "Result" component, it says that $-\dfrac{1}{(2-3x)^2}=-\dfrac{1}{(3x-2)^2}$, which I have explained above. –  Sujaan Kunalan Nov 19 '13 at 18:06

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