Evaluating Limits
Lesson 3.3
Let’s Review
Limit – Given f (x), a limit, L, exists if f (x) becomes
arbitrarily close to a single number L as x
approaches c from either side
x 2  3x  2
f ( x) 
x2
lim f ( x) 
x2
1
Let’s Focus
Properties of limits:
1) Limit of a constant is the constant
lim b  b
lim 3  3
x 9
x c
Direct Substitution Limits (i.e. if you substitute the constant
for the variable, evaluate, and nothing weird happens, then you’re good)
2) Limit of x as x approaches n is n
lim x  9
lim x  c
x c
x 9
3) Limit of x to an integer power n is xn.
lim x  c
n
xc
n
lim x 2  92  81
x 9
Stay Focused!
More Direct Substitution Limits:
Given b,cR, nZ, and lim f ( x)  L and lim g ( x)  K exist, then…
x c
x c
4)
Constant factors can be factored out of a limit
lim b[ f ( x)]  bL
x c
5)
lim 2 x 2  2[lim x 2 ]  2(16)  32
x4
x4
Limits of sums of functions are the sum of their limits
lim [ f ( x)  g ( x)]  L  K
x c
lim [ x 2  3] 
x1
lim x 2  lim 3  1  3  4
x 1
6)
x 1
Limits of products of functions are the product of their
limits
lim [ f ( x) g ( x)]  LK
x c
lim [(2 x 2 )(3x)] 
x1
lim (2 x 2 ) lim (3x)  (2)(3)  6
x 1
x1
Just…A…Little…More…Focus…
Yet More Direct Substitution Limits
7) The limit of a polynomial function p(x) as x
approaches c is p(c).
lim p ( x)  p (c)
x c
lim 4 x 2  2 x  36  6  42
x3
8) The limit of a rational function r(x) as x approaches c
is r(c), as long as the quotient of r(x) at c is not zero.
lim r ( x)  r (c) 
x c
p (c )
, q (c )  0
q (c )
x 2  x  2 10 5
lim


2
x 3
x 1
4
Too…Much…Focusing…ARGH!
Still Yet More Direct Substitution Limits
9) The limit of a radical function n√x as x approaches c
is n√c (if n√c exists).
lim
x c
n
x c
n
lim 3 x 2  4
x 8
10) The limit of a composite function f (g(x)) is f (L).
lim
x 0
x2  4  4  2
(Finally) Getting Down to Business
Limits of Functions Identical at All But One Point
g ( x)  a , then
Given f (x) = g(x) for all x except x=c, if lim
x c
lim f ( x)  a
x c
x2  x  2
lim
?
x 1
x 1
x 2  x  2 ( x  1)( x  2)
f ( x) 

x 1
( x  1)
 ( x  2)  g ( x)
Since f (x) = g(x) at all points except x = 1, and
f ( x)  3
lim g ( x)  (1  2)  3 , then… lim
x 1
x 1
Another Technique: Rationalizing
lim
x 0
x 1 1
?
x
Recall…
1) (a + b)(a – b) = a2 – b2
f ( x) 
x  1  1  x  1  1 
 x 1 1
x


 x  1 1 

 

 x( x  1  1) 
g ( x) 
2) (√x)2 = x


 x  1 2  12 


 x( x  1  1) 




x




x
(
x

1

1
)


1
, and lim
x 0
x 1 1
3) x(1) = x

1
x 1 1
1
 1 , so lim f ( x)  1
2
2
x 0
x 1 1
Another Quick Word…
The following theorem is required to be instructed as part of the AP exam. All of two
classwork questions are included for practice from the textbook, one of which is very
challenging to do.
The study of this theorem can be a very helpful when completing Calculus proofs, but
that is beyond the scope of our study. The unlikely possibility that you will encounter
the need to use this theorem results in the “light exposure” approach that will follow.
Man, these Calculus lessons are getting long. And why does the instructor continue
to use these “Quick Word” slides? They only make it longer.
Thank you. And now back to your regularly-scheduled lesson.
The Squeeze Theorem
Squeeze Theorem – If a function f (x) is bound (“squeezed”)
between functions g(x) and h(x), so h(x) ≤ f(x) ≤ g(x) for all x in
g ( x)  L  lim h( x) , then
an open interval containing c, and if lim
x c
x c
lim f ( x)  L
x c
sin( x)
lim
?
x 0
x
lim h( x)  1
x 0
lim g ( x )  1
x 0
g ( x )  f ( x )  h( x )
sin( x)
 lim
1
x 0
x
Practice
Find each limit.
1) lim 3 15  x 2
x 4
2)
2 x 2  5x  3
lim
x 3
x3
Classwork
Pg. 224 (5-21 odd; 27-47 odd)