When demonstrating that
lim
x
→
2
2
x
+
5
=
9
\displaystyle \lim_{{{x}\to{2}}}{2}{x}+{5}={9}
x
→
2
lim
2
x
+
5
=
9
with
ϵ
=
0.3
\displaystyle \epsilon={0.3}
ϵ
=
0.3
, which of the following
δ
\displaystyle \delta
δ
-values suffices?
δ
=
\displaystyle \delta=
δ
=
0.3
δ
=
\displaystyle \delta=
δ
=
0.05
δ
=
\displaystyle \delta=
δ
=
0.0225
δ
=
\displaystyle \delta=
δ
=
0.15
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