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I am having difficult proving that the limit $$\lim_{x\to 1} \frac{x-2}{x^4-1}$$ does not exist using the epsilon-delta definition.

Clearly this must be true since the function is unbounded near 1, but I'm having difficult formalizing this.

Any help to point me in the right direction would be appreciated.

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    $\begingroup$ You can, for example, find $x$ arbitrarily close to $1$ such that ${x-2 \over x^4-1}$ has values $\ge 1$ and values $\le -1$. This would not be possible if a limit exists. $\endgroup$ – copper.hat Oct 15 '17 at 20:56
  • $\begingroup$ For $|x-1|<0.5$, notice that$$\left|\frac{x-2}{x^4-1}\right|=\frac{|x-2|}{|x-1|\cdot|1+x|\cdot|1+x^2|}\ge\frac{2-1.5}{|x-1|\cdot(1+2)\cdot(1+2^2)}$$ $\endgroup$ – Simply Beautiful Art Oct 15 '17 at 21:06
  • $\begingroup$ @copper.hat Thank you for replying. I was thinking this as well, and one thought I had earlier was to prove that the left hand limit at 1 must (exist) and equal the right hand limit at 1 for the overall limit to exist, but this is impossible since one side is positive and the other negative. I'm still a little hazy on how to formulate this in terms of epsilon-delta, however, could you help me there? $\endgroup$ – user18097 Oct 15 '17 at 21:07
  • $\begingroup$ You can show that $\lim\limits_{x\to1^+}f(x)=-\infty$ and $\lim\limits_{x\to1^-}f(x)=+\infty$. $\endgroup$ – Aretino Oct 15 '17 at 21:13
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The fraction can be written as $$ \frac{1}{x-1}\frac{x-2}{(x+1)(x^2+1)} $$ Taking $0<x<1$ we have $$ 1-x>0,\quad 2-x>1,\quad x+1<2,\quad x^2+1<2 $$ so $$ \frac{1}{x-1}\frac{x-2}{(x+1)(x^2+1)}= \frac{1}{1-x}\frac{2-x}{(x+1)(x^2+1)}> \frac{1}{2(1-x)} $$ and, for $M>0$ we have $$ \frac{1}{2(1-x)}>M $$ as soon as $$ 1-x<\frac{1}{2M} $$ that is, $$ x>1-\frac{1}{2M} $$ Thus the function is unbounded in a left neighborhood of $1$. This shows the limit cannot be finite.

The limit can be neither $\infty$ nor $-\infty$, because the function is positive in a left neighborhood of $1$ and negative in a right neighborhood of $1$.

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Let $f(x) = {x-2 \over x^4 -1 }$ and note that $\lim_{x \uparrow 1} f(x) = \infty$ and $\lim_{x \downarrow 1} f(x) = -\infty$.

Hence there are $a_n <1$, $a_n \to 1$ such that $f(a_n) \ge 1$ for all $n$, and similarly, there are $b_n >1$, $b_n \to 1$ such that $f(b_n) \le -1$ for all $n$.

To show that the limit does not exist, we must show that for any $L$, there exists some $\epsilon >0$ such that for any $\delta>0$ there is some $x \in B(1,\delta)\setminus \{1\}$ such that $|f(x)-L| \ge \epsilon$.

Suppose $L \ge 0$ first. Set $\epsilon = 1$ and pick $\delta >0$. Then there is some $b_n$ such that $|b_n-1| < \delta$ and $|f(b_n)-L| \ge 1 = \epsilon$.

Now suppose $L <0$ and use the $a_n$ in a similar way to get $|f(a_n)-L| \ge 1 = \epsilon$.

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$\lim_{x\to 1} \frac{x-2}{x^4-1} $

Let $x = 1+y$. then

$\begin{array}\\ \lim_{x\to 1} \frac{x-2}{x^4-1} &=\lim_{y\to 0} \frac{(1+y)-2}{(1+y)^4-1}\\ &=\lim_{y\to 0} \frac{y-1}{1+4y+6y^2+4y^3+y^4-1}\\ &=\lim_{y\to 0} \frac{y-1}{4y+6y^2+4y^3+y^4}\\ &=\lim_{y\to 0} \frac{y-1}{y(4+6y+4y^2+y^3)}\\ &=\lim_{y\to 0} \frac{-1}{4y}\\ \end{array} $

and this limit does not exist (and has different signs for $y > 0$ abd $y < 0$).

Note that if the numerator was $x-1$ instead of $x-2$, then the limit would be $\dfrac14$ since the quotient would be $\frac{y}{y(4+6y+4y^2+y^3)} =\frac{y}{4y} =\frac14 $.

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