Computing Limits

Intuitively, we say that limx→ c f(x) = L if f is defined near (but not necessarily at) c and f(x) approaches L as x approaches c.

If we let x approach c from the left side only, we write limx→ c- f(x) since x is approaching c from smaller values. Similarly, for x approaching c from the right, we write limx→ c+ f(x). The two-sided limit limx→ c f(x) exists if and only if both of these one-sided limits exist and they are equal.

An Intuitive Example

Consider the graph of a function f(x) shown below.

Evaluate each of the following. Select (click) each one to check your reasoning.

•   f(2)
          •   f(-5)
          •  
lim
x→ -3 
 		f(x)
•  
lim
x→2+ 
 		f(x)
          •  
lim
x→-5+ 
 		f(x)
          •  
lim
x→ - ∞ 
 		f(x)
•  
lim
x→2- 
 		f(x)
          •  
lim
x→-5- 
 		f(x)
          •  
lim
x→ ∞ 
 		f(x)
•  
lim
x→ 2 
 		f(x)
          •  
lim
x→ -5 
 		f(x)
          •  
lim
x→c 
 		f(x)  for  c ≠ -5, -3, 2.

Definition of the Limit

More rigorously, let f be defined at all x in an open interval containing c, except possibly at c itself.

Then


lim
x→ c 
 f(x) = L
if and only if for each ε > 0, there exists a δ > 0 such that
if  0 < |x-c| < δ then  |f(x)-L| < ε.

In words, limx→ c f(x) = L if and only if by taking x close enough to c (but different from c: see c = −5 in the preceding example) we can get f(x) arbitrarily close to L.

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Properties of the Limit

Each of the following properties is proven using the rigorous definition of the limit. Let lim stand for limx→ c, limx→ c+, or limx→ c-. Assume lim f(x) and lim g(x) both exist.

  • (Uniqueness)   If lim f(x) = L1 and lim f(x) = L2, then L1 = L2.
  • (Addition)   lim [f(x)+g(x)] = lim f(x) +lim g(x).
  • (Scalar multiplication)   lim [αf(x)] = αlim f(x).
  • (Multiplication)   lim [f(x)g(x)] = lim f(x) · lim g(x).
  • (Division)   lim [f(x)/ g(x)] = lim f(x) / lim g(x), provided lim g(x) ≠ 0.
  • (Powers)   lim [f(x)]n = [lim f(x)]n for any positive integer n.

In practice, much of the time we can ``reason out'' the value of a limit without explicitly using the ε-δ definition.

Examples

  • limx→ 2 √(x2+12) = 4 since the function f(x) = √(x2+12) is continuous at x = 2 and f(2) = 4.
  • limx→ ∞ 1/x = 0 since as x increases, 1/x gets arbitrarily close to 0.
  • limx→ 0+ ln|x| tends to -∞ and so does not exist since as x decreases to 0, ln|x| gets arbitrarily large in magnitude and negative.
  • limx→ 3 [(x2-9)/( x-3)] = 6 even though f(x) = (x2-9)/( x-3) is undefined at x = 3 since (x2-9)/( x-3) = x+3 and limx→ 3x+3 = 6.

What about something like limx→ 0 [sin x / x]? When we cannot easily ``reason out'' the value of a limit, we can often use numerical methods or L'Hôpital's Rule to determine the value of the limit. Can you convince yourself that limx→ 0 [sin x / x] = 1?


Key Concepts [index]

  • Let the function f be defined at all x in an open interval containing c, except possibly at c itself.

    Then


    lim
    x→ c 
    		 f(x) = L
    if and only if for each ε > 0, there exists a δ > 0 such that
    if  0 < |x-c | < δ then  |f(x)-L| < ε.

    In words, the limit of f(x) as x approaches c is L if and only if by taking x close enough to c (but different from c) we can get f(x) arbitrarily close to L.