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I'm trying to prove that a limit does not exists for the following expression:

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

I might have missed something trivial here, but my attempt to prove that this limit goes to infinity, i.e. does not exist, goes as follows (am I wrong or right here?):

First, $$\frac{x^2-1}{x^2-2x+1} = \frac{x+1}{x-1}.$$ Next, let us first consider the left limit and therefore only consider say $[0,1)$. As $x\to 1^-$ we see that $x+1 = 2$. Moreover, let $x-1 = \frac{a}{b}$ for some $a,b$, we know that $-1 \leq \frac{a}{b} < 0$ as long as we move from 0 and approaches 1. This means that as $x \to 1^-$ then $$\frac{x+1}{x-1} = \frac{2}{\frac{a}{b}} = \frac{2b}{a} = \infty$$ since $\frac{a}{b}$ gets infinitely small (hence $b$ grows infinitely big).

Is this correct way to prove this? If not, how should one do it without showing a graph

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  • $\begingroup$ You have to use L'Hospital. And then you get it doesn't exist. $\endgroup$
    – dylan7
    Aug 30 '14 at 16:24
  • $\begingroup$ @dylan7 how would you use l'hospital? There is no indeterminate form, direct substitution yields $\frac{2}{0}$, which is undefined. $\endgroup$
    – Varun Iyer
    Aug 30 '14 at 16:25
  • $\begingroup$ Yes, it's correct. However, as long as you get $\frac{x+1}{x-1}$, you know that the limit as $x\rightarrow1$ does not exist, as the limit is $\neq0$ in numerator, and $0$ in denominator. $\endgroup$ Aug 30 '14 at 16:26
  • $\begingroup$ @Varun lyer Before you perform L'Hospital you get $\frac {0}{0}$ on thr original expression. Once you do it you get $\frac {2}{0} $. Then conclude it doesn't exist. $\endgroup$
    – dylan7
    Aug 30 '14 at 16:27
  • $\begingroup$ @dylan7 ahh my bad I understand now $\endgroup$
    – Varun Iyer
    Aug 30 '14 at 16:28
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$$\lim_{x \to 1} \frac{x^2-1}{x^2-2x+1}=\lim_{x \to 1}\frac{(x-1)(x+1)}{(x-1)(x-1)}=\lim_{x \to 1}\frac{x+1}{x-1}=\infty$$

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    $\begingroup$ Be careful. $$\lim_{x \to 1^+}\left[\frac{x+1}{x-1}\right]=\infty;$$ ${}$ $$\lim_{x \to 1^-}\left[\frac{x+1}{x-1}\right]=-\infty.$$ ${}$ Therefore the limit does not exist. $\endgroup$
    – beep-boop
    Aug 30 '14 at 16:35
  • $\begingroup$ @alexqwx: this is what I had in mind in my attempt, but can't we more formally show the two cases? I mean, the right limit is shown just by "reason" that $x-1$ constitutes a really small number, which further implies that the whole fraction gets really big, thus goes to infinity. But can we show this formally (as I tried with my $\frac{a}{b}$ thing..) ? $\endgroup$
    – hsalin
    Aug 30 '14 at 19:03

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