2.5 Evaluating Limits
Algebraically
Mr. Peltier
Indeterminate Forms
• For continuous functions, we use substitution
to find the limit
• What about discontinuous functions at the
discontinuities?
• Basic limit laws do not apply to limits of
indeterminate forms, such as:
– 0/0, ∞/∞, or ∞ - ∞
Indeterminate Forms
• So what do we do if f(x) is indeterminate at
x = c?
• Try to algebraically simplify the function so
that it is now continuous at x = c
Using Algebra
• EX: Evaluate
x2  x  2
x lim 2
x2
Substitution gives us 0/0,
so we try some factoring
x 2  x  2  ( x  1)( x  2)
( x  1)( x  2)
x lim 2
x2
x
lim 2 ( x  1)  3
Time to cancel!
Using Trigonometry
• EX: Evaluate
x
lim 
2
tan x
sec x
sin x
cos x
1
sec x 
cos x
tan x 
Substitution gives us ∞/∞,
so we try some identities
x lim 
2
x
sin x
cos x
1
cos x
lim  sin x
2
 x lim 
2
1
Time to cancel!
sin x cos x

cos x 1
Using Algebra
• EX: Evaluate x lim 4
x 2
x4
Substitution gives us 0/0, so
we try using the conjugate
(this is a method that
sometimes works with roots)
x
x
lim 4
lim 4

x 2 x 2

x4
x 2


x4
Time to cancel!
x lim 4
x  4 x  2
1
1

x lim 4
x 2
4






x 2 x 2
x  4 x  2
FOIL the
top
Using Algebra
2 
 1
 2 
• EX: Evaluate x lim 1 
 x 1 x 1 
Substitution gives us ∞ - ∞,
so we add the fractions
(need common denom)
 1

2
x lim 1 
 x  1  ( x  1)( x  1) 


 1x  1

2
x lim 1 
 x  1x  1  ( x  1)( x  1) 


 x 1 2 

 Time to cancel!
x 1
  x lim 1 

 x lim 1 
 x  1x  1 
 x  1x  1 
 1 
1
 
 x lim 1 
2
 x  1 
Assignment
Page
#1, 7, 9, 11,
17, 19, 26,
27, 31, 35,
39, 46, 48