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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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  • $\begingroup$ Is the $h \to \infty$ intentional? Or is the question about why $(3x-2)^2=(2-3x)^2$ $\endgroup$
    – Bonnaduck
    Nov 19, 2013 at 0:34
  • $\begingroup$ 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. $\endgroup$
    – Dorus
    Nov 19, 2013 at 14:08
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    $\begingroup$ 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. $\endgroup$
    – Bonnaduck
    Nov 19, 2013 at 17:17
  • $\begingroup$ Thanks, that pretty much answers my question. $\endgroup$
    – Dorus
    Nov 20, 2013 at 0:00

2 Answers 2

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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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  • $\begingroup$ 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? $\endgroup$
    – Dorus
    Nov 19, 2013 at 10:29
  • $\begingroup$ 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? $\endgroup$ Nov 19, 2013 at 17:59
  • $\begingroup$ 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. $\endgroup$ Nov 19, 2013 at 18:06
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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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  • $\begingroup$ 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] $\endgroup$
    – Dorus
    Nov 19, 2013 at 14:13

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