Limit Theory (page 8)
Thm 10/ Uniqueness Theorem If lim f ( x)  L then L is unique.
(Indirect) Proof/ Assume also that lim f ( x )  L'  L . (Argue to a contradiction.)
The idea here is that there can’t be 2 limits since the y-values have to fall within a small
neighborhood about y = L. So if we take as an error margin an epsilon ½ the distance
between L' & L , if the y-values land all in one neighborhood, they can’t be in the other.
Let   12 L  L' however, since both limits exist there exist two deltas…
1  f ( x )  L   and  2  f ( x )  L '   , now let   min 1,  2 
 L  f ( x )  f ( x )  L '  2 ( since |a – b| = |b – a| )
and  L  f ( x)   ( f ( x)  L ')  L  f ( x)  f ( x)  L ' (since |a + b|  |a| + |b| ).
Using the transitive property for inequalities, we get: L  L'  2 .
However, from above we also have: L  L'  2 .
The contradiction leads to the conclusion that L  L' . QED
(Direct) Proof/ Let lim f ( x )  L and lim f ( x )  M
x a
x a
The idea here is to show that if there are two limits, they have to equal each other.
So for each epsilon (or for an even smaller epsilon, say 2 , there exist
1  0  x  a  1  f ( x )  L  2 and  2  0  x  a   2  f ( x )  M  2
Adding the two inequalities, we obtain: M  f ( x )  f ( x )  L  2  2  
Using the Triangle Inequality Theorem ( |a + b|  |a| + |b| ) that:
M  f ( x)   f ( x)  L   M  f ( x)  f ( x)  L
we get: M  f ( x)   f ( x)  L    or
 M  L  0  
Hence, we’ve shown that lim  M  L   0 which means
x a
lim M  lim L  0 (The limit of a sum equals the sum of the limits.)
xa
x a
Finally M = L (The limit of a constant function equals that constant.)
Hence, the limit value is unique.
Limit Theory (page 9)
Thm 11/ Limits of nth roots Theorem
(A) For n odd, if lim f ( x)  L, then lim n f ( x)  n lim f ( x)  n L
xa
xa
xa
(B1) For n even & L > 0, if lim f ( x)  L, then lim n f ( x)  n lim f ( x)  n L
xa
xa
xa
(B2) For n even & L = 0 and f ( x)  0  x in a deleted neighborhood about x  a ,
then lim n f ( x )  n lim f ( x)  n L  0
xa
xa
Thm 12/ If f is differentiable at x = a, then f is continuous at x = a.
f ( x )  f (a )
Proof/ f differentiable implies: lim
 f '(a ) .
xa
x a
f ( x )  f (a )
Now consider f ( x )  f (a ) 
x  a
xa
Since the limit of the last factor is just ‘0’, that is: lim  x  a   0 , we have
x a
f ( x)  f (a )
f ( x)  f (a)
 lim  x  a   lim
0  0
xa
xa
x a
x a
x a
x a
 lim f ( x )  f (a ) which is the definition for continuity at a point, x = a.
lim  f ( x )  f (a )   lim
x a
Thm 13/ Derivative of a Constant Function – If f ( x)  k , then f '( x)  0  x R .
f ( x )  f (a )
k k
Proof/ lim
 lim
0
xa
x a
x a x  a
Thm 14/ Another Slide Theorem – If f is differentiable, then
d
d
 k  f ( x)  k  f ( x)
dx
dx
Proof/ This proof uses the ‘Slide Theorem for Limits’.
k  f ( x  h)  k  f ( x)
f ( x  h)  f ( x)
d
lim
 k lim
 k  f ( x) 
h
h
dx
h0
h0
Thm 15/ If f ( x)  x, then f '( x)  1 .
f ( x)  f (a )
xa
Proof/ lim
 lim
1
xa
x a
x a x  a
Limit Theory (page 10)
 
d n
x  n  x n 1 for any positive integer, n.
dx
f ( x  x )  f ( x )
y
Proof/ Given y  x n , we will construct the difference quotient, x =
x
Using f ( x )  x n and f ( x  x )  ( x  x )n and the Binomial Theorem, we get:
Thereom 16a/ Power Rule
y  ( x  x )n  x n 
2
r
n ( n 1) n  2
x
x  ...  n Cr x n  r x  ... 
2
2
r
n
n ( n 1) n 2
x
x  ...  n Cr x n r x  ...  x
2
x n  nx n 1x 
So
or
y
x
y
x

nx n 1x 
 
 
 
 
 x n 
xn
 
x
2
r 1
n 1
n ( n 1)
 nx n 1  2 x n 1  x   ...  n Cr x n r  x   ...   x 
y
xa x
Finally, we have: lim
 nx n 1 , QED
Thm 16b/ Extending the power rule to include n = 0: For y  x 0 ,
d 0
x  0  x 1 
dx
0
x
d 0
x  0 x  0
dx
 0  x  0 (Also recall 00 is undefined.)
Thm 16c/ Extending the power rule to include negative integers, n < 0:
d n
x  n  x n 1
dx
Proof/ Since ‘n’ is negative if and only if ‘-n’ is positive…
d u
v  du  u  dv
d n d  1 

x   n  . Now use the quotient rule
v
dx
dx
dx  x 
v2
 


x  n  0  1  nx  n 1
 x n 
2
  nxn1  nxn1  x2n  nxn1 QED
x 2n
Thm 16d/ Extending the power rule to include rational exponents, r .
 
d r
x  rx r 1
dx
There will be restrictions on the variable x.
Proof/ Let r  mn in reduced form with m & n integers.
Also let n be positive so that if r < 0, m < 0.
d r
d  mn  d m n1 1 m n1 1
x  x 
x
n x
 mx m 1 (The last step used the chain rule.)
dx
dx   dx
 
 
m m m1
m
xn
n
 
 rx r 1 QED