Epsilon-delta limit does not exist 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.
 A: 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
$|f(a_n)-L| \ge 1 = \epsilon$.
A: $\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
$.
A: 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 $1$.
