Solución
Falso;
lim
x
→
6
f
(
x
)
\mathop {\lim }\limits_{x \to 6} f\left( x \right)
x
→
6
lim
f
(
x
)
DNE ya que
lim
x
→
6
−
f
(
x
)
=
2
≠
5
=
lim
x
→
6
+
f
(
x
)
\mathop {\lim }\limits_{x \to {6^ - }} f\left( x \right) = 2 \ne 5 = \mathop {\lim }\limits_{x \to {6^ + }} f\left( x \right)
x
→
6
−
lim
f
(
x
)
=
2
=
5
=
x
→
6
+
lim
f
(
x
)