10. Limits involving infinity

10. Limits involving infinity
It is known from the limit rules for fundamental arithmetic operations (+,-,×,÷) that if
two functions have finite limits at a (finite or infinite) point x0 , that is, they are
convergent, the limit of their sum, product, etc. will be equal to the sum, product, etc. of
their limits (except in case of a quotient with zero denominator). In other words, the
operation of finding the limit is interchangeable with the fundamental arithmetic
operations.
In the table below we consider those cases, when one of the two functions (or both)
tends to infinity (is proper divergent), or the denominator function tends to 0. In such
situations two cases are possible: either the limit of the expression of fundamental
arithmetic can be given in advance (definite limit expression), or it cannot be given in
advance in lack of additional information, and the limit can be anything or even may not
exist (indefinite limit expression), in the latter type of cases the unknown limits are
symbolized by question marks.
Theorem (limit rules involving infinity)
ASSUMPTION
STATEMENT
1
SYMBOLIC
NOTATION
lim ( f ( x )  g ( x ))  
x x0
lim ( f ( x )  g ( x ))  ?
x x0

 ?
If lim f ( x)   and lim g ( x)  , then
x  x0
x  x0
lim f ( x )  g ( x )  
x x0
lim
x x0
f ( x)
?
g ( x)
  

?

3
3
2
lim ( f ( x )  g ( x ))  
c
lim ( f ( x )  g ( x ))  
c    
 , if c  0
lim f ( x)  g ( x)   , if c  0,
c    , if c  0
x x0
x x0
If lim f ( x)  c and lim g ( x)  , then
x  x0
x  x0
(c real constant)
x  x0
?, if c  0
lim
x x0
f ( x)
0
g ( x)
If lim f ( x)   or lim f ( x)  c  0,
x  x0
x  x0
and lim g ( x)  0, then
x  x0
(0+: g(x)0 and g(x)>0 around x0, that
lim
f ( x)

g ( x)
lim
f ( x)
?
g ( x)
x x0
c    , if c  0
0  ?
3
c
0


 ,
0
c
 , if c  0
0
is, it tends to 0 from the positive side)
If lim f ( x)  0 and lim g ( x)  0, then
x  x0
1
x  x0
x x0
0
?
0
3
All statements of the theorem remain valid, if xx0 if replaced by the one sided limits xx0, xx0+, x or
x  (so in the assumption as in the corresponding statement).
2
The equations in the symbolic notation must not be considered as ordinary equations (e.g, you must not subtract
from both sides , etc.). The left side of the equation symbolizes the type of the limit expression, the right side
gives the limit in definite cases or a question mark symbolizes uncertainty in the indefinite cases.
3
The indefinite expressions of type / and 0/0 can be handled by means of l’Hospital’s rule. The type 0 . can
be reduced to type 0/0 via the transformation f(x).g(x)= f(x)/(1/g(x)) The difference of type  can be reduced to
a quotient type by bringing to a common denominator or using other transformations.
Limits of composite functions
Theorem. If lim g ( x )  u0 and lim f (u )  L , the
x  x0
uu0
lim f ( g ( x))  lim f ( g ( x))  lim f (u)  lim f ( x)  L .
x x0
g ( x )u0
uu0
xu0
If, in addition, u0 is finite and f(x) is continuous at u0, then
lim f ( g ( x ))  f ( lim g ( x ))  f (u0 )
x x0
x x0
Remark. The theorem also holds true for one-sided limits including limits at infinity
(x0,. u0,.L can be  or ). Without „lim”notation we can also write: xx0 
u=g(x)u0  f(u)=f(g(x))L .
Example: lim cot(e  x )  lim cot(u)  , because: x  u=e–x0+ 
x 
u 0 
cot (u)=cot (e–x)  .
L’Hospital’s rule
Theorem. If lim f ( x )  lim g ( x )  0 (type 0/0) or lim f ( x )  lim g ( x )   (type
x x0
x x0
x x0
x x0
/), and both functions are differentiable around x0, then lim
x x
0
f ( x)
f ( x )
,
 lim
g ( x ) x x0 g ( x )
provided the latter limit exists.
Remark. The theorem also holds true for one-sided limits including limits at infinities
(“around x0” means then a one-sided environment). In addition to the type / it can
also be applied to the types /, /() and /(), as well.
Important: Always check the conditions of applicability before using the rule’s
formula! If not used for the proper type of expression, false results are to be expected.
Examples:
sin x
cos x
 lim
 cos 0  1 .
x0 x
x0
1
a) type 0/0: lim
b) type 0.(), change into type /, use l’Hospital’s, then simplify and substitute:
lim x ln x  lim
x 0 
x 0 
ln x
1
x
 lim
x 0 
1
x
1
x2

 lim ( x)  0 .
x 0 
c) type , reduce to quotient, first use composite’s and then l’Hospital’s rule:
lim (ln(1  e x )  x )  lim (ln(1  e x )  ln e x )  lim ln
x
x
x

1  ex
1  ex


ln
lim
 x e x
ex



ex
  ln  lim x

 x e



  ln 1  0


ln(f(x))g(x) g(x) ln
Power expressions: use the exponential transscription (f(x))g(x)=e
=e
f(x)
and find the limit of the formula in the exponent. For example, utilizing the result of
the previous example b) (substituting u=x ln x0) we can find this limit:
lim x x  lim e ln x  lim e x ln x  lim e u  e 0  1 .
x
x 0 
x 0 
x 0 
u 0
Comparing the rates of convergence or proper divergence
Definition 1. Assume that lim f ( x ) lim g ( x ) 0 . Then we use the following terminology:
x  x0
x x0
0 : f(x) tends to 0 faster (in order of magnitude) than g(x), as xx0
if lim
x  x0
f ( x)

g ( x)
c0 : f(x) and g(x) tend to 0 in the same order of magnitude
(c=1: equally fast), if xx0
 : f(x) tends to 0 slower (in order of magnitude) than g(x), as xx0
Definition 2. Assume that lim f ( x ) lim g ( x )  . Then we use the following
x  x0
x  x0
terminology:
 : f(x) tends to  faster (in order of magnitude) than g(x), as xx0
if lim
x x0
c0 : f(x) and g(x) tend to  in the same order of magnitude
f ( x)

g ( x)
(c=1: equally fast), if xx0
0: f(x) tends to  slower (in order of magnitude) than g(x), as xx0
It is exactly l’Hospital’s rule to use for such rate comparisons, because the assumptions
are the same. For example, by the above example a), f (x) = sin x and g(x) = x tend
equally fast to 0 as x0, because their quotient tends to c=1.
PROBLEM 1
LIMITS OF RATIONAL FRACTIONAL FUNCTIONS AT A FINITE POINT
Determine the following limits:
lim
a)
x 2
x2  4x  4
x2  4
x2  2x  1
2
b) x1 x  1
lim
x2  2x  2
3x 2
c) x0
3x 3  2 x 2  x
x2
d) x0
lim
lim
PROBLEM 2
LIMITS OF RATIONAL FRACTIONAL FUNCTIONS AT INFINITY
Determine the following limits:

x2 x3 
a) lim 1  x 
 
x
2
6 

b) lim (4 x 2  2 x 4 )
3x 3  2 x 2  x
d) lim
x
4x  3
6x5  4x3  2x
e) lim 6
x x  x 4  x 2  1
x
x 
x2  2x  2
3x 2
2 x
f) lim   
x   x
2
9
2007 x 2006 2006 x
x 
x  2008
(2 x 2  x ) 2
x  1  3x 2  5 x 4
g) lim
PROBLEM 3
c) lim
h) lim
LIMITS OF ELEMENTARY FUNCTIONS
Give the limits of the elementary functions (sketch the graph and observe the limit):
lim 1
a)
x 3
f)
x64
lim
lim x 2
lim 2
x
b)
x 
g)
x64
lim
3
x
c)
x 
h)
x 
lim 3 x
lim x 3
d)
i)
x
e) lim
x 0 
lim 2 x
x
j)
lim e x
x 
x
lim e x
k)
x
p)
x 0
l) lim 10 x
m) lim lg x
x 0
x 0
q)
2
lim ln x
x
o)
lim ln x
x
lim cos x
lim sin x
lim log 1 x
n)
x 
r)
x0
s) lim tan x

t) lim cot x

2
2
x
x
lim lg x
u)
x10
v) lim tan x
PROBLEM 4

w) lim tan x

x 0
x
x) lim cot x
x 0
4
DEFINITE LIMIT EXPRESSIONS; COMPOSITE FUNCTIONS
Determine the following limits:
a) lim ( x 2  e x )
x
d) lim
x 2
tan x
2x  
log 0,5 x
g) lim
2
x 0 
cos x
b) lim ( x 2  e x )
x
x
x 1 ln x
c) lim (ln x  cot x)
x 0 
e) lim
f) lim x 3 ( 12 ) x
h) lim x 3 ln x
i) lim
x
x
1
x  ln x
1
 1
 
j) lim 
x 0  sin x
x


Determine the limits of the following composite functions:
a) lim ln(1  x 2 )
b) lim sin
c) lim x e
d) lim x e
e) lim cot e x
f) lim tan
x 1
x 0 
g) lim
x
x 
1 

 ln x 
x 

x 0 



 1  cos x 
x 0
h) lim ln 
1  cos x
x 
x 
 i)
x
  1
lim lg(cos x )
x0
j) lim
x
x 0 
x
PROBLEM 5

INDEFINITE LIMIT EXPRESSIONS: L’HOSPITAL’S RULE
Determine the following limits:
1  e 2 x
x 0
x
a) lim
b) lim
lg x
x10
c) lim
d) lim xe x
e) lim ( x   ) cot x
f) lim
x 
x 
x 
x 1
x 
sin x
ln x
sin 2 x
1sin x cos x
e x  4x2
x  e x  2 x
j) lim
x 
sin( cos x )
x 0
sin 2 x
l) lim ln x  ln(1  x)
k) lim
log 2 x
x 1
i) lim
x0
e x  e
tg x
m) lim
1  cos x
x 0 x sin x
h) lim x ln x 2
g) lim
n) lim
x 1
x 2
p) lim x 1  x
ln sin x
cos 2 x
q) lim (cos x) x
x 0
x 1
o) lim x x
x 
 1
1 
r) lim 


x 1 ln x
x 1 

2
x 0 
PROBLEM 6
Find out the limits of the two functions (sketch their graphs) and compare the rate of
their convergence (or proper divergence)!
a) x and (1  cos x), as x0
b) ln x and x, as x
c) e x and x, as x
d) x and
e) log x and cot x, as x0
f) 2 x and x 2 , as x
g) ln x and ( x1) 2 , as x1
h) cot x and 1x , as x0
i) tan x and 2 x1 , as x 2
j) e  x and x 1 , as x
k) cot πx and ln 21x , as x 12
l) ln x and 3 x-1, as x1
1
2
PROBLEM 7
, as x0
l
ln x
FINDING LIMITS FOR THE EXAMINATION OF FUNCTIONS
Where is the function continuous (domain)? Determine the (possibly one-sided) limits
of the function at the end-points of its domain! (Attention: l’Hospital’s rule is not
always applicable!)
a) x 2 ( x  2) 2
x2
1  x2
b) 2 x 32 x 2  x 1
c)
1
x
f)
g) xe x
h) xe  x
2
i)
j) e1/ x
k) e 1/ x
2
d)
m)
ln x
x
p) x 2  ln x 2
e) x 
n)
1
x  2x  3
2
x  x2
ex
x
l) x ln x
x  1
o) ln 

 x 
x
ln x
q) cos( e x )
2
r)
x
x
RESULTS
1. a) 0 b) 2 c)  d) does not exist (left: , right: )
2. a)  b)  c) 1/3 d)  e) 0 f)  g) 4/5 h) 
3. a) 1 b) 2 c)  d)  e) 0 f) does not exist (not defined) g) 4 h)  i) 0 j)  k) 0
l) 1 m)  n) does not exist (not defined) o)  p)  q) does not exist (oscillating) r)
1
s) does not exist (left: , right: ) t) 0 u) 1 v) 0 w) 1 x) 
4. a)  b)  c)  d)  e) does not exist (left: , right: ) f)  g)  h)  i) 0 j) 
5. a)  b) 0 c) 0 d)  e)  f)  g)  h) 0 i) 0 j) 0
6. a) 2 b) 0 c)  d) 0 e) 1 f) 2 g) 1 h) 0 i) 1/2 j) e k) /2 l) 0 m) 0 n) 1/2
o) 1 p) e q) e–1/2 r) 1/2
7. a) 0; x slower b) ; ln x slower c) ; e x faster d) 0; x faster
e) ; log1/2 x slower f) ; 2 x faster g) 0; ln x slower
h) only one-sided limits exist: from the left , from the right ; equally fast
i) only one-sided limits exist: from the left , from the right ; equal order of magnitude
(c=2)
j) 0; e –x faster k) 0; equal order of magnitude (c=/2) l) 0; ln x faster
8. a) D =(,), Lim: both at  and at : =  b) D =(,), Lim: at : = , at : = 
c) D =(,3)(3,1)(1,), Lim: both at  and at : = 0; at 3 left: = , right: = ; at 1
oppositely
d) D =(,), Lim: both at  and at : =1
e) D =(,0)(0,), Lim: at : = , at : =; at 0: = , at 0+: = 
f) D =[0,1], Lim: both at 0+ and at 1: = 0 g) D =(,), Lim: at : = 0, at : = 
h) D =(,), Lim: both at  and at : = 0
i) D =(,0)(0,), Lim: at : = 0, at : = ; at 0: = , at 0+: = 
j) D =(,0)(0,), Lim: both at  and at : = 1; at 0: = 0, at 0+: = 
k) D =(,0)(0,), Lim: both at  and at : = 1; at 0: = 0 l) D =(0, ), Lim: at 0+: = 0; at :
=
m) D =(0,), Lim: at 0+: = ; at : = 0
n) D =(0,1)(1,), Lim: at 0+: = 0; at : = ; at 1: = , at 1+: = 
o) D =(,1)(0,), Lim: both at  and at : = 0; at 1: = ; at 0+: = 
p) D =(,0)(0,), Lim: both at , at  and at 0: = 
q) D =(,), Lim: both at  and at : = 1 r) D =(0, ), Lim: at 0+: = 0; at : = 1.

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