Is limit of function -1/0 ok?
Emily Wong 
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 $-∞$?
$\endgroup$ 45 Answers
$\begingroup$$$\frac{x^2-2x}{x^2-2x+1}=\frac{x(x-2)}{(x-1)^2}\xrightarrow[x\to1]{}-\infty$$
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.
$\endgroup$ 4 $\begingroup$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 $\infty$.
$\endgroup$ 5 $\begingroup$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 $$ \lim_{x→1}\frac{x^2 - 2x}{x^2 -2x +1}=-\infty $$
$\endgroup$ $\begingroup$Notice, $$\lim_{x\to 1}\frac{x^2-2x}{x^2-2x+1}$$ $$\lim_{x\to 1}\frac{(x^2-2x+1)-1}{x^2-2x+1}$$ $$=\lim_{x\to 1}\left(1-\frac{1}{(x-1)^2}\right)\longrightarrow \color{red}{-\infty}$$
$\endgroup$ $\begingroup$$$\lim\limits_{x\to 1}\frac{x^2-2x}{x^2-2x+1}$$
$$=\lim\limits_{x\to 1}(x^2-2x)\lim\limits_{x\to 1}\frac{1}{x^2-2x+1}$$
$$=-1\cdot\lim\limits_{x\to 1}\frac{1}{\underbrace{x^2-2x+1}_{\to 0+}}=\color{red}{-\infty}$$
$\endgroup$
