Proofs
Theorems involving Limits

Definition of a Limit
Let f be a function defined on an open interval containing c (except possibly at c) and let
L be a real number. The statement lim f ( x)  L means that for each 𝜀 > 0 ∃ a 𝛿 > 0 ∋ if
0 | x  c |  then | f ( x )  L |  .
x c
(if you chose a value close enough to c, you can get the value of your function as close
as you like to the limit)

Basic Limits (2.1)
1. lim b  b
(the limit of a constant is itself constant)
x c
The proof of this is obvious since the value of x does not affect the value of b.
2. lim x  c
x c
To prove this, let ε = δ, then if |x-c| < δ, then |x-c| < ε.

3. lim x n  lim x
x c
x c

n
 cn
This can be proved for the more general case below.

Properties of Limits (2.2)
Let lim f ( x)  L and lim g ( x)  K
x c
x c
4. lim bf ( x)  b lim f ( x)   bL
(scalar multiple)
xc
 xc

This proof is a straightforward application of the definition, by factoring out the
constant from the absolute values.
5. lim  f ( x)  g ( x)  lim f ( x)  lim g ( x)  L  K
x c
x c
x c
(sum & difference)
This proof requires the use of the Triangle Inequality: |x + y| ≤ |x| + |y|.
Proof: Choose ε > 0. Then ε/2 > 0, ∴ δ1 > 0 exists ∋ 0 | x  c | 1 implies
| f ( x )  L |

2
and δ2 > 0 exists ∋ 0 | x  c |  2 implies | g ( x )  K |

2
. Let δ be
the smaller of δ1 and dδ2. Then 0 | x  c |  implies both | f ( x )  L |
| g ( x )  K |

2

2
and
. The Triangle inequality tells us that
| [ f ( x )  L]  [ g ( x )  K ] || [ f ( x )  g ( x )]  ( L  K ) || f ( x )  L |  | g ( x )  K |

2


2

Thus lim[ f ( x)  g ( x)]  L  K  lim f ( x)  lim g ( x) . Difference case done
x c
x c
x c
similarly.
Quod erat demonstrandum.



6. lim  f ( x) g ( x)  lim f ( x) lim g ( x)  LK
xc
xc
x c
(product)
Proof: Because the limit of f(x) is L and the limit of g(x) is K, we have that
lim[ f ( x )  L]  0 and lim[ g ( x)  K ]  0 . Let 0 < ε < 1. Then ∃ δ > 0 ∋ if
x c
x c
0 | x  c |  , then | f ( x )  L  0 |  and | g ( x )  K  0 |  implies that
| [ f ( x )  L][ g ( x )  K ]  0 || f ( x)  L || g ( x)  K |    . So
lim[ f ( x)  L][ g ( x)  K ]  0 , and furthermore, by property 4, lim Lg ( x)  LK
x c
x c
and lim Kf ( x)  KL . Write that
x c
f ( x ) g ( x )  [ f ( x )  L][ g ( x )  K ]  [ Lg ( x )  Kg ( x )]  LK . (You can check this by
distributing.) Taking the limit of the above expression: lim f ( x ) g ( x ) 
x c
lim[ f ( x)  L][ g ( x)  K ]  lim Lg ( x)  lim Kg ( x)  lim LK  0  LK  LK  LK  LK
x c
x c
x c
x c
Quod erat demonstrandum.
7. lim
x c
f ( x) L
f ( x ) lim
 x c
 ,K  0
g ( x ) lim g ( x ) K
(quotient)
x c
1
1
 . The rest will follow directly from
x c g ( x )
K
Property 5. Let ε > 0. Since lim g ( x )  K ∃ δ1 > 0 ∋ if 0 | x  c | 1 then
Proof: It is enough to prove that lim
x c
|K |
| g ( x )  K |
. (the reason for choosing this value will be clear later.) This
2
|K |
implies that | K || g ( x )  [| K |  g ( x )] || g ( x ) |  || K |  g ( x) || g ( x) | 
. And
2
|K |
1
2
| g ( x ) | ,
since
. Similarly, ∃ δ2 > 0 ∋ 0 | x  c |  2 then

2
g ( x) | K |
| K |2
 . (The reason for choosing this value will be clear later.) Let δ
2
be the smaller of δ1 and δ2. Then for 0 | x  c |  you have
| g ( x )  K |
1
1
K  g ( x)
1
1
1
2 | K |2




| K  g ( x ) |


   . We
g ( x) K
g ( x)K
| K | g ( x)
|K | |K | 2
obtained this expression by finding a common denominator, factoring out the
denominator and making substitutions from the previous steps. This
1
1
lim
 .
x c g ( x )
K
Quod erat demonstrandum.
n
n
8. lim  f ( x)  lim f ( x)   Ln
(powers)
x c
 xc

The proof for this is a straight-forward application of Property 6 and
mathematical induction (at least for the integer case). The base case can be for
f ( x )  f ( x )  [ f ( x )]2 , and then the second case assume true for [ f ( x )]n , and
prove for [ f ( x)]n1  [ f ( x)]n  f ( x) .

Limits involving Radicals (2.4)
9. lim n x  n c
xc
Proof: The following proof works well n ∈
positive. For any ε > 0, find a δ > 0 ∋
n
, and c > 0 for the case where n is
x  c   whenever | x  c |  . This is
n
equivalent to saying that   n x  n c   and   x  c   . If we assume
that   n c implies that 0  n c    n c . Let δ be the smaller of c 
and

n

n
c 

 c 


n
c 

n

c xc
the nth root.
n
c 

n
 c . That gives us   x  c   . Replace expressions for δ.
n
c    x c 

n

n

n

n
c 
c 

n

n
 c . Distribute on the left.
 c . Add c.

c    n x  n c   . Subtract

n

n
c 
n
c .   n x  n c   .
x
n
c 
 . Take
n
Quod erat demonstrandum.


10. lim f ( g ( x))  f lim g ( x)  f ( L)
xc
xc
Proof: For a given ε > 0, we need to find a δ > 0 ∋ | f ( g ( x ))  f ( L) |  for
0 | x  c |  . Because the limit of f(x) as x→L is f(L), know that ∃ δ1 > 0 ∋
| f (u )  f ( L) |  for | u  L | 1 . Because the limit of g(x) as x→c is L, you know
∃ δ > 0 ∋ | g ( x)  L | 1 for 0 | x  c |  . Letting u=g(x), we have the proof.
Quod erat demonstrandum.
11. limsin( x)  sin(c)
x c
The proofs of the transcendental functions and trig functions are left to an
advanced calculus course.
12. limcos( x)  cos(c)
x c
13. lim tan( x)  tan(c)
x c
14. limcot( x)  cot( c)
x c
15. limsec( x)  sec(c)
x c
16. limcsc( x)  csc(c)
x c
17. lim a x  a c (a  0)
x c
18. limln( x)  ln(c)
x c

Functions that Agree at all but one point (2.7)
Let c be a real number and let f(x)=g(x) for all x≠c in an open interval containing c. If the
limit of g(x) as x approaches c exists, then the limit of f(x) also exists and
lim f ( x)  lim g ( x) .
x c
x c
This theorem can be extended to any function that differs from another by a finite
number of points.
Proof: Let L be the limit of g(x) as x goes to c. Then, for each ε > 0 ∃ a δ > 0 ∋ f(x)=g(x) in
the open interval (c – δ,c)and (c,c + δ) and | g ( x )  L |  when 0 | x  c |  . Because
f(x)=g(x) for all x in the open interval other than x=c, it follows that | f ( x )  L |  when
0 | x  c |  . This the limit of f(x) as x goes to x is also L.
Quod erat demonstrandum.

The Squeeze Theorem (2.8)
If h(x) ≤ f(x) ≤ g(x)  x in an open interval containing c, except possibly at c itself, and if
lim h( x )  L  lim g ( x ) then lim f ( x ) exists and is also equal to L.
x c
x c
x c
Proof: For ε > 0 ∃ δ1 > 0 and δ2 > 0 ∋ | h( x )  L |  for 0 | x  c | 1 and,
| g ( x )  L |  for 0 | x  c |  2 . Because h(x) ≤ f(x) ≤ g(x)  x in an open interval
containing c, except possibly c itself, ∃ δ3 > 0 ∋ h(x) ≤ f(x) ≤ g(x) for 0 | x  c |  3 . Let δ
be the smallest of the three. Then if 0 | x  c |  , it follows that | h( x )  L |  and
| g ( x )  L |  which implies that   h( x )  L   and   g ( x )  L   . Add to get
L    h ( x ) and g ( x )  L   taking just the left side of the first inequality and the
right side of the second. Since h(x) ≤ f(x) ≤ g(x) we can put this together with the above
inequalities to obtain L    h( x )  f ( x )  g ( x )  L   or L    f ( x )  L   . ∴
lim f ( x)  L .
x c
Quod erat demonstrandum.

Special Limits (2.9)
sin( x )
1
19. lim
x 0
x
The textbook presents a geometric proof for the expression. You can also
demonstrate this to yourself numerically. Other types of proofs are left to an
advanced calculus course.
20. lim
x 0
1  cos( x )
0
x
21. lim(1  x)
x 0

1
x
e
Vertical Asymptotes (2.14)
Let f be continuous on an open interval containing c. If f(c)≠0, g(c)=0 and ∃ an open
interval containing c ∋ g(x)≠0  x≠c in the interval, then the graph of this function given
f ( x)
by h( x ) 
has a vertical asymptote at x=c.
g ( x)
Proof: Consider the case for which f(c) > 0 (the case for f(c) < 0 is similar). ∃ b>c ∋
c<x<b implies g(x)>0. Then for M > 0 (any number, the upper limit), choose δ1 ∋ 0 < x – c
f (c)
3 f ( c)
 f ( x) 
< δ1 implies that
and δ2 ∋ 0 < x – c < δ2 implies that
2
2
f (c)
0  g ( x) 
. Let δ be the smaller of δ1 and δ2. Then it follows that 0 < x – c < δ
2M
f ( x)
f ( x ) f (c)  2 M 
implies that

 M . So it follows that lim
  and the line x=c


x c g ( x )
g ( x)
2  f (c) 
is a vertical asymptote of the graph h.
Quod erat demonstrandum.

Definition of a Horizontal Asymptote
The line y=L is a horizontal asymptote of the graph of f if lim f ( x )  L or lim f ( x)  L .
x 
x 

Limits at Infinity (4.10)
c
 0 and
x  x r
If r is a positive rational number and c is any real number, then lim
c
 0 . (the second limit is only valid if x r is defined when x < 0. ) lim e x  0 and
r
x 
x  x
x
lim e  0 .
lim
x 
1
 0 . For ε > 0, let M = 1/ε. Then, for x > M, you have
x  x
1
1
1
x  M       0   . So, by the definition of a limit at infinity, you can

x
x
conclude that the limit of 1/x as x→∞ is 0. Now, using this result and letting r = m/n,
you can write the following
Proof: Begin by proving lim
  1 m 


c
c
1
1
lim r  lim m  c  lim  n    c  lim n   c  n lim   c
x  x
x 
x 
x  x
x
 x  x  
x n



The second part of the theorem is similar.
m
m
 0
n
m
 0.
Quod erat demonstrandum.

Existence of a limit (2.10)
Let f be a function and let c and L be real numbers. The limit of f(x) as x approaches c is L
iff lim f ( x )  L and lim f ( x)  L .
x c
x c
Properties of continuity dependent on limit properties. So all these properties can likewise be applied
to similar continuity issues.