In formally proving that
lim
x
→
6
(
x
2
+
x
)
=
42
\displaystyle \lim_{{{x}\to{6}}}\ {\left({x}^{{2}}+{x}\right)}={42}
x
→
6
lim
(
x
2
+
x
)
=
42
, let
ϵ
>
0
\displaystyle \epsilon>{0}
ϵ
>
0
be arbitrary. Choose
δ
=
\displaystyle \delta\ =\
δ
=
min
(
ϵ
m
,
1
)
\displaystyle {\left(\frac{\epsilon}{{m}}\ ,\ {1}\right)}
(
m
ϵ
,
1
)
. Determine the smallest value of
m
\displaystyle {m}
m
that would satisfy the proof.
m
=
\displaystyle {m}=\
m
=
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