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 Applied calculus topic summary: limits and continuity

Tools: Function Evaluator & Grapher | Excel Grapher

Subtopics: Limits | Estimating Limits Numerically | Estimating Limits Geometrically | Computing Limits Algebraically | Continuous Functions

Limits

We write

 limx→a f(x) = L
or  f(x) → L as x → a
to mean that f(x) approaches the number L as x approaches (but is not equal to) a from both sides.

A more precise way of phrasing the definition is that we can make f(x) be as close to L as we like by making x be sufficiently close to a.

We write

 limx→a+ f(x) = L
or  f(x) → L as x → a+
and
 limx→a- f(x) = L
or  f(x) → L as x → a-
to mean that f(x) → L as x approaches a from the right or left, respectively. For limxa f(x) to exist, the left and right limits must both exist and must be equal.

We write

 limx→+∞ f(x) = L
and
 limx→-∞ f(x) = L
to mean that f(x) → L as x gets arbitrarily large or becomes a negative number with arbitrarily large magnitude, respectively.
Examples

1. As x → 3, the quantity 3x2-4x+2 approaches 17, and hence

 limx→3 (3x2-4x+2) = 27
Notice this is just the value of the function at x = 3 (see "Algebraic Approach" below).

2. Notice that the function

 x2 - 9x - 3
is not defined at x = 3. However, for other values of x, it simplifies to
 x2 - 9x - 3 = (x - 3)(x + 3)x - 3 = x + 3,
and, as x → 3, this quantity approaches 6. Therefore
 x2 - 9x - 3 → 6 as x → 3,

or  limx→3 x2 - 9x - 3 = 6

There are many more examples in the on-line tutorials on limits.

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Estimating Limits Numerically
 To analyze a limit of the form limx→a f(x) or limx→±∞ f(x) numerically:
• Make a table of values for f(x) using values of x that approach a closely from either side.
• If the limit exists, then the values of f(x) will approach the limit as x approaches a from both sides.
• The more accurately you wish to evaluate this limit, the closer to a you will need to choose the values of x.
• For a limit as x → +∞, use positive values of x getting larger and larger.
• For a limit as x-∞, use negative values of x getting larger and larger in magnitude.
Examples
 1. To estimate limx→3 x2 - 9x - 3 , constuct a table with values of x close to 3 on either side:

x approaching 3 from the left

x approaching 3 from the right
x
2.9
2.99
2.999
2.9999
 f(x) = x2 - 9x - 3
5.9
5.99
5.999
5.9999
 3
 3.0001 3.001 3.01 3.1 6.0001 6.001 6.01 6.1

Since the values of f(x) appear to be approaching 6 as x approaches 3 from either side, we estimate that the limit is 6.

 2. To estimate limx→ +∞ x2 - x + 12x2 - 3 , constuct a table with values of x approaching +∞:

x approaching +∞   →
x
10
100
1000
10,000
 f(x) = x2 - x + 12x2 - 3
0.461929
0.495124
0.499501
0.49995
 +∞

Since the values of f(x) appear to be approaching 0.5 as x approaches 3 from either side, we estimate that the limit is 0.5.

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Estimating Limits Geometrically
 To analyze a limit of the form limx→a f(x) or limx→±∞ f(x) geometrically:
• Draw the graph of f(x) either by hand or using technology, such as a graphing calculator.
• If you want to calculate a limit as xa for a real number a, position your pencil point (or the graphing calculator "trace" cursor) on a point of the graph to the left of x = a.
• Move the point along the graph toward x = a from the left and read the y-coordinate as you go. The value the y-coordinate approaches (if any) is then the limit
 limx→a- f(x)
• Repeat the above two steps, but this time starting from a point on the graph to the right of x = a, and approach x = a along the graph from the right. The value the y-coordinate approaches (if any) is then
 limx→a+ f(x)
• If the left and right limits both exist and have the same value L, then
 limx→a+ f(x)
exists and equals L.
• If you want to calculate a limit as x → +∞, position your pencil point (or the graphing calculator "trace" cursor) on a point near the right edge of the graph, and move the pencil along the graph to the right, estimating the y-coordinate as you go. The value the y-coordinate approaches (if any) is then the limit
 limx→+∞+ f(x)
For x-∞, start toward the left edge, and move your pencil toward the left.
Example

The following picture shows the graph of f(x). First, we calculate the limit of f(x) as x → 0 from the left:

 limx→0- f(x)

Thus,
 limx→0- f(x)
= -1

Next, we calculate the limit of f(x) as x → 0 from the right:
 limx→0+ f(x)

Thus,
 limx→0+ f(x)
= 3
Since the left- and right limits disgaree, we conclude that
 limx→0 f(x)
does not exist.

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Computing Limits Algebraically:
Limits as xa
 To compute a limit of the form limx→a f(x) algebraically:
1. Check to see whether f is a closed form function. These are functions specified by a single formula involving constants, powers of x, radicals, exponentials and logarithms, combined using arithmetic operations and composition of functions.
1. If a is in the domain of f, then limxa f(x) = f(a).
2. If a is not in the domain of f, but f(x) can be reduced by simplification to a function with a in its domain, then (a) applies to the reduced form of the function.
3. If a is not in the domain of f, and you cannot simplify the function as in (b), then simplify as much as possible, and evaluate the limit by the numerical approach, or, if you know its graph, by the graphical approach.
2. If f is not closed-form, and a is a point at which the definition of f changes, compute the left limit and right limit seperately, and check whether they agree.

Limits as x±∞
 To compute a limit of the form limx→±∞ f(x) algebraically:

If x is approaching ±∞, then check to see whether f(x) is a ratio of polynomials. If it is, then you can ignore all but the highest powers of x in the numerator and denominator. This simpler function will have the same limit as f.

Examples
 1 Consider the limit limx→1 x2 - 9x - 3 .
 Notice that the function f(x) = x2 - 9x - 3 is closed form, and a = 1 is in its domain.
Therefore, the limit is obained by substituting x = 1 (point (a) opposite):
 limx→1 x2 - 9x - 3 = 1 - 91 - 3 = 4
 2 Now consider limx→3 x3 - 9x - 3 .
This time, a = 3, which is not in the domain of f, and so we need to first simplify f(x) to reduce it to a function that does have 3 in its domain:
 limx→3 x2 - 9x - 3 = limx→3 (x - 3)(x + 3)x - 3 = limx→3 x + 3.
Now, x = 3 is in the domain of f, and so we find the limit by putting x = 3:
 limx→3 x2 - 9x - 3 = limx→3 x + 3 = 3 + 3 = 6.
 3 Consider limx→+∞ x3 + x2 - 92x3 - x - 3 .
Here f(x) is a ratio of polynomials, so we ignore everything except the highest power of x in the numerator and denominator:
 limx→+∞ x3 + x2 - 92x3 - x - 3
=  limx→+∞ x3 + x2 - 92x3 - x - 3
=  limx→+∞ x32x3
=  limx→+∞ 12 = 12
(Cancel the x2)

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Continuous Functions

A function f is continuous at a if limxa f(x) exists, and is equal to f(a).

The function f is said to be continuous on its domain if it is continuous at each point in its domain. The algebraic approach to limits above is based on the fact that any closed form function is continuous on its domain.

Examples

The function f(x) = 3x2-4x+2 is a closed form function, and hence continuous at every point in its domain (all real numbers).

The function

g(x)= 4x2+1x - 3
is also a closed form function, and hence continuous on its domain (all real numbers excluding 3).

On the other hand, the function

h(x)=  -1 if -4 ≤ x < -1 x if -1 ≤ x ≤ 1 x2-1 if 1 < x ≤ 2
is not a closed-form function, and is in fact discontinuous at x = 1. (See the tutorial on continuity.)

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Last Updated: May 2007