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<h1>Solución</h1>
<p>Supongamo que $L$ es candidato a límite. Elegimos $\epsilon = 1/2.$
<p class="apar">Sea $\delta &gt 0.$</p>
<p> Se cumple $L \ge 0$ o $L &lt 0$. Si $L \ge 0$, entonces sea $x = −\delta / 2$.
Así,
$$\left| {x - 0} \right| = \left| { - {\delta  \over 2} - 0} \right| = {\delta  \over 2} &lt \delta $$
y
$$\left| {{{\left| { - {\delta  \over 2}} \right|} \over { - {\delta  \over 2}}} - L} \right| = \left| { - 1 - L} \right| = L + 1 \ge 1 &gt {1 \over 2} = \varepsilon $$
Por otro lado, si $L &lt 0$,
entonces sea $x = \delta / 2$.
Así,
$$\left| {x - 0} \right| = \left| {{\delta  \over 2} - 0} \right| = {\delta  \over 2} &lt \delta $$
y
$$\left| {{{\left| {{\delta  \over 2}} \right|} \over {{\delta  \over 2}}} - L} \right| = \left| {1 - L} \right| = \left| L \right| + 1 \ge 1{\rm{  &gt }}{1 \over 2} = \varepsilon $$
Por tanto, para cualquier valor de $L$, $$\mathop {\lim }\limits_{x \to 0} {{\left| x \right|} \over x} \ne L$$


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