Download 2.2. Computing Limits

2.2. Computing Limits
• First we will obtain the limits of some simple functions.
• Then based on these, we will find limits of more complicated
functions
Basic Limits
2.2.1 Theorem. Let a and k be real numbers.
(a) lim k  k
x a
(b) lim x  a
x a
1
(c) lim  
x0 x
1
(d ) lim  
x 0 x
In Part (c), let
f ( x)  k is a constant function, we have
lim(kg ( x))  lim k lim g ( x)  k lim g ( x)
x a
x a
x a
xa
and similarly for one-sided limits.
Note:
• Although parts (a)-(c) of Theorem 2.2.2 are stated for two functions,
the results hold for any finite number of functions.
• Moreover, the various parts of the theorem can be used in
combination to reformulate expressions involving limits.
For example,
lim[ f ( x)  g ( x)  2h( x)]  lim f ( x)  lim g ( x)  2lim h( x)
x a
x a
x a
x a
lim[ f ( x) g ( x)h( x)]  (lim f ( x))(lim g ( x))(lim h( x))
x a
x a
lim[ f ( x)]3  (lim f ( x))3
x a
x a
lim[ f ( x)]n  (lim f ( x)) n
x a
x a
lim x 3  (lim x)3  a 3
x a
x a
x a
x a
Limits of polynomials as x  a
For example, Find lim( x3  3x  2).
x4
Solution : lim( x3  3 x  2)  lim x 3  lim3 x  lim 2
x 4
x 4
x 4
x 4
= lim x 3  3lim x  lim 2
x 4
x 4
=43  3(4)  2
=64+12-2=74
x 4
2.2.3 Theorem
For any polynomial
p( x)  c0  c1 x  ...  cn x n
and an real number a,
lim p( x)  c0  c1a  ...  cn a n  p (a )
x a
6
4
2
For example, Find lim( x  3 x  1) .
x 4
Solution: The function involved is a polynomial. So
lim( x6  3x4 1)2  (46  3(4)4 1)2  23648769
x4
Limits of Rational Functions as x  a
2
3
x
1
Example, Find lim
x1 x  2
2
lim(3 x 2  1) 3(1) 2  1 2
3
x

1
Solution: lim
 x1


x1 x  2
lim( x  2)
1 2
3
x1
2 x
x5 ( x  5)( x  2)
Example: Find lim
Solution: when x=5, (x-5)(x+2) is a zero, but x-2 is not a zero. So
2 x
does not exist.
x5 ( x  5)( x  2)
lim
Etc.
When p(x)/q(x) is a rational function for which p (a) =0 and q(a)=0, the
numerator and denominator must have one or more common factors
of x – a.
In this case, the limit of p(x)/q(x) as x  a can be found by canceling
all common factors of x – a first. Here are some examples…
For example,
x2 1
Find xlim
 1 x  1
Solution: Since 1 is a zero of both the numerator and the denominator,
they share a common factor of x-1. The limit can be obtained as
follows:
x2 1
( x  1)( x  1)
lim
 lim
 lim ( x  1)  2
x 1 x  1
x 1
x 1
x 1
Example: Find
x2  x  6
(a) lim
x 2
x2
Solution: The numerator and the denominator both have a zero at x=2,
so there is common factor of x-2. Then
x2  x  6
( x  2)( x  3)
lim
 lim
 lim( x  3)  5
x2
x

2
x2
x2
x2
2
x
Find (b)lim  4 x  3
x3 x 2  6 x  9
Solution: The numerator and the denominator both have a zero at x=3,
so there is common factor of x-3. Then
x2  4 x  3
( x  3)( x  1)
( x  1)
(b) lim 2
 lim
 lim
x3 x  6 x  9
x3 ( x  3)( x  3)
x3 ( x  3)
Since lim( x  1)  2,lim( x  3)  0
x3
x3
x2  4 x  3
( x  1)
So lim 2
 lim
does not exist.
x3 x  6 x  9
x3 ( x  3)
Limits of Piecewise-defined functions
For functions that are defined piecewise, a two-sided limit at a point
where the formula changes is best obtained by first finding the onesided limits at that point.
Example:
t 2 , t  0
Let g(t )  
t  2, t  0
Find
(a) lim g (t ),(b) lim g (t ),(c) lim g (t )
x0
Solution:
x0
x0
(a) lim g (t )  lim(
t  2)  0  2  2

x0
x0
(b) lim g (t )  lim t 2  (0) 2  0
x0
x0
(c) lim g (t )  lim g (t ), thus lim g (t ) does not exist.
x0
x0
x0