y
=
e
x
Here we show that
y
=
f
(
x
) =
e
x
is continuous at
x
0
for any real number
x
0
. To do
so, we must show that
lim
x
→
x
0
f
(
x
) =
lim
x
→
x
0
e
x
is equal to
f
(
x
0
) =
e
x
0
.
Let’s deFne Δ
x
by
x
=
x
0
+ Δ
x
(so
x
→
x
0
is equivalent to Δ
x
→
0), and use the
fact that
e
x
0
+Δ
x
=
e
x
0
e
Δ
x
.
Then what we are left to show is that
lim
x
→
x
0
h
e
x
0
e
Δ
x
i
=
e
x
0
.
But
x
0
is held constant as
x
→
x
0
, so the factor
e
x
0
can be taken outside the limit
(by the constantmultiple limit theorem), leaving us with the problem of showing that
e
x
0
±
lim
x
→
x
0
h
e
Δ
x
i²
=
e
x
0
or, equivalently, that
lim
x
→
x
0
h
e
Δ
x
i
= 1
,
(1)
where the
lim
x
→
x
0
is equivalent to
lim
Δ
x
→
0
.
We will prove this last relation using
the algebraic (
²

δ
) formulation of the limit problem discussed in class, and in the
textbook.
PROOF OF EQUATION (1)
We Frst set the condition of closeness of the function,
e
Δ
x
, to the claimed limit, 1,
using
² >
0 as the measure of this degree of closeness:
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Summer '10
 ZETO
 Calculus, Continuity, Derivative, e∆x, constantmultiple limit theorem

Click to edit the document details