<s>[INST] 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.
 [/INST] $\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
$.
</s>