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Tutorial 1 Comments
Q1
f
(
x
) =
x

x

. To nd
f
0
(0)
, we have by de nition
f
0
(0) = lim
x
→
0
f
(
x
)

f
(0)
x
= lim
x
→
0
x

x

if the limit
EXISTS.
Now consider the function
g
(
x
) =
x

x

. This function is continuous for all
x
∈
R
.
Recall that for a function
g
(
x
)
which is continuous for all
x
∈
R
, by de nition,
lim
x
→
c
g
(
x
) =
g
(
c
)
,
for all
c
∈
R
.
Therefore,
f
0
(0) = lim
x
→
0
x

x

= lim
x
→
0
g
(
x
) =
g
(0) = 0
.
Recall too that the reason for which the function
h
(
x
) =

x

is not di erentiable at
x
= 0
is that
f
0
(0) = lim
x
→
0
f
(
x
)

f
(0)
x
= lim
x
→
0

x

x
DOES NOT EXIST
!
We can see it in two ways :
1) The function

x

x
is
NOT
continuous. (There is a jump at x=0 since

x

x
=

1
for
x <
0
and

x

x
= 1
for
x >
0
.
2) As stated in the lecture notes, the lefthand and right hand limits are not equal, and
thus the limit does not exist. (we can see this as also an intepretation of the fact that

x

x
is
not continuous)
Q2
When encountered with problems such as nding the derivative of a product of functions
(more than 3 say),
f
(
x
) =
g
(
x
)
h
(
x
)
j
(
x
)
k
(
x
)
,
an alternative way to approach the problem is to take ln on both sides of the equation so
that the product becomes a sum.
ln
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This note was uploaded on 11/30/2011 for the course EEE 1001 taught by Professor Phoon during the Spring '11 term at National University of Singapore.
 Spring '11
 phoon

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