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A quick question, i'm determining the limit of this function:

$$\lim_{x→1}\frac{x^2 - 2x}{x^2 -2x +1}$$

When I divide numerator and denominator by $x^2$ and fill in $1$, I get $-1/0$. This is an illegal form right? Or does it indicate it is going to $∞$ or $-∞$?

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    $\begingroup$ the numerator does not tend to zero so your limit will diverge. Since it does not tend to the same value then you can not apply the L'Hopital's test. $\endgroup$
    – Chinny84
    Commented Mar 4, 2016 at 12:12
  • $\begingroup$ @5xum Opps I meant it can not. Updated the error! $\endgroup$
    – Chinny84
    Commented Mar 4, 2016 at 12:15
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    $\begingroup$ Incidentally, there's no need to divide the numerator and denominator by $x^{2}$. (If you were taking a limit as $x \to \infty$ or $x \to -\infty$, however, that operation would be appropriate.) $\endgroup$ Commented Mar 4, 2016 at 12:27
  • $\begingroup$ my hint: if you get $-1/0$ then it is $-\infty$ that is yur correct answer $\endgroup$ Commented Mar 4, 2016 at 13:58

5 Answers 5

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$$\frac{x^2-2x}{x^2-2x+1}=\frac{x(x-2)}{(x-1)^2}\xrightarrow[x\to1]{}-\infty$$

since $\;x(x-2)\to-1\;$ and $\;(x-1)^2\to 0^+\;$ (meaning: approximates to zero from the positive side), so your limit is negative infinity.

Some may define this as "the limit doesn't exist", but I think it is more accurate to say "the limit doesn't exist finitely" and/or "the limit exists in the wide sense of the word", " the function diverges to $\;-\infty\;$", or something similar.

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    $\begingroup$ The terminology I was taught (in highschool, but repeated in university courses as well) was to say that the limite exists and is infinite. The "limit doesn't exist" statement was left for the cases when no value, finite or infinite, would satisfy the limit definition (like $\lim_{x \to +\infty} \sin(x)$). $\endgroup$
    – Bakuriu
    Commented Mar 4, 2016 at 13:26
  • $\begingroup$ @Bakuriu I was alternatively said "the limit exists, wheter finite or infinite", or in case it is infinite then we can also say "it diverges to infinity"...it all depends on the lecturer and/or author, I guess. $\endgroup$
    – DonAntonio
    Commented Mar 4, 2016 at 14:10
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    $\begingroup$ The precise statements would be "The limit doesn't exist in $\mathbb R$" or "The limit exists and is infinity in the extended reals" - or, if you want to make all of your rational functions continuous (which is kind of fun) - "The limit exists and is infinite in the one point compactification of the reals/the Riemann sphere." Whether or not a limit exists is clear cut if you know the space you're working in. $\endgroup$ Commented Mar 4, 2016 at 15:51
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    $\begingroup$ What's with the double spaced formatting? $\endgroup$
    – Kyle Kanos
    Commented Mar 4, 2016 at 17:58
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There's certainly nothing illegal about the form. It may be a nonsensical expression, but so far, no law has been acepted that prohibits it.

If you get $\frac{-1}{0}$ as the result, then the limit can either not exist or be equal to $-\infty$ or $\infty$. For example, the limit $$\lim_{x\to 0}\frac{1}{x}$$

does not exist, while the limit$$\lim_{x\to0}\frac1{x^2}$$ is equal to $\infty$.

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  • $\begingroup$ @Nikunj Thanks. Strangely enough, this is the second time today I read the word "illegal" on this site. A couple hours ago, I aparently gave an "illegal" downvote to a question :P $\endgroup$
    – 5xum
    Commented Mar 4, 2016 at 12:45
  • $\begingroup$ I think the OP might have meant illegal for L'Hospital's rule – or not satisfying its assumptions (which is true). Because that's where one classifies limits as being of type 0/0 or ∞/∞ and the like. $\endgroup$
    – The Vee
    Commented Mar 4, 2016 at 13:25
  • $\begingroup$ "illegal" is a very widely known technical term that means "not allowed" and does not only refer to judicial laws. If your computer crashes, and especially if it gives you a blue-screen, it might say "Illegal exception" or the like, meaning that something was executed that was not allowed and hence the whole process was terminated. $\endgroup$
    – user21820
    Commented Jul 30, 2016 at 6:42
  • $\begingroup$ @user21820 Yes, I know. It was a light hearted joke to start the answer. $\endgroup$
    – 5xum
    Commented Aug 1, 2016 at 6:34
  • $\begingroup$ Haha okay. I seriously thought you didn't know the technical meaning, and certainly some mathematics students don't. $\endgroup$
    – user21820
    Commented Aug 1, 2016 at 6:37
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For $x\ne1$, you have $x^2-2x+1>0$; on the other hand, the numerator is negative at $1$, so it is negative in a neighborhood of $1$. Since the denominator has limit $0$, you can conclude that $$ \lim_{x→1}\frac{x^2 - 2x}{x^2 -2x +1}=-\infty $$

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Notice, $$\lim_{x\to 1}\frac{x^2-2x}{x^2-2x+1}$$ $$\lim_{x\to 1}\frac{(x^2-2x+1)-1}{x^2-2x+1}$$ $$=\lim_{x\to 1}\left(1-\frac{1}{(x-1)^2}\right)\longrightarrow \color{red}{-\infty}$$

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$$\lim\limits_{x\to 1}\frac{x^2-2x}{x^2-2x+1}$$

$$=\lim\limits_{x\to 1}(x^2-2x)\lim\limits_{x\to 1}\frac{1}{x^2-2x+1}$$

$$=-1\cdot\lim\limits_{x\to 1}\frac{1}{\underbrace{x^2-2x+1}_{\to 0+}}=\color{red}{-\infty}$$

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