ALGEBRAIC LIMITS

ALGEBRAIC LIMITS

 A LIMIT IS A Y-VALUE

When we find a limit for an algebraic function, we find the value that y or f (x) APPROACHES as the x coordinate APPROACHES or gets extremely close to a given value.

NOTICE THE EMPHASIS ON THE WORD APPROACHES!!!

If y = 3x + 1 then

means that the y coordinate approaches 7 as x approaches 2.

In this case, y equals 7 when x = 2 because this linear function is defined at x = 2, so 2 is an element of the domain for y = 3x + 1.

Read "The Y's the Limit" -- article explaining limits.

 If f (a) exists then,

For this one, all we did was substitute x = 2, but when we get an indeterminate form by substitution, there is no precise y-value for the given x because the expression is undefined. To find the limit as x approaches a for an expression or function which is not defined at x = a, there are various techniques.

Factor, Reduce, Substitute

Example: find

since becomes at x = 2, it is an indeterminate form,
so we find the value that y approaches as x approaches 2.

The graph of this function will have a hole -- a missing point at .

Since 2 is not an element of the domain, will be missing from the range.

Note the form -- once we substitute x = a and evaluate, we no longer write " ".

Here's one where factoring is done by division.

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Rationalize, Reduce, Substitute

Find . If we set x = 4, we get , so we rationalize the numerator.

As x approaches 4, the expression will approach . Since x cannot = 4, the expression will never be equal to .

This one can be done using factor, reduce, substitute if we think of x as .

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Absolute Value Expressions

To find when , we first do an analysis of the function on either
side of x = 4, since that's where the numerator changes sign. The only difference between the numerator and the denominator is the sign (+ or – ). The numerical value will be the same.

When x < 4, f (x) = – 1 (numerator is > 0 (abs. val. > 0) but denominator is < 0 since x < 4.)

When x > 4, f (x) = +1 (numerator is > 0 (abs. val. > 0) but denominator is > 0, since x > 4.)

In such a case, we discuss one-sided limits.

So, for , we define the left-side and right-side limits.(see diagram)

..... and .....

Since .... , we say that does not exist.

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One-Sided Limits

Notation: A left limit is denoted -- notice the minus sign above the a in the subscript. A right limit is denoted -- notice the plus sign above the a in the subscript.

 If f (x) is defined on the open interval (c, a) then the left-side limit means that f (x) gets very close to L 1, as x gets very close to a, with x < a (approaching from the left of a). If f (x) is defined on the open interval (c, a), then the right-side limit means that f (x) gets very close to L 2 as x gets very close to c, with x > c (approaching from the right of c).

(diagram)

 If f (x) is defined over an open interval including "a " except possibly at "a " itself then exists and = L if and only if

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Functions Defined on Restricted Domains

For the function , the domain is [ – 3, 3] so we can only consider one-sided limits for the extremes of this function. and

since the domain of of this function is , x can only approach – 3 from above it (right-side limit) and x can only approach +3 from below it (left-side limit).

Piecewise Defined Functions

 A PIECEWISE DEFINED FUNCTION is one in which the function is defined differently on specific intervals or pieces of the domain.

In this example the "pieces" are left of and right of x = 1. On the left of and at x = 1, we plot the linear function y = 2x – 1. Right of x = 1, we plot the parabola y = x².

..... and .....
, therefore exists and = 1.

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However, if the function had been defined as:

the left limit , but the right limit .

The left limit is not equal to the right limit, so does not exist.

Practice

Find the limits:

A/

 1.a) b) c) d) 2.a) b) c) 3.a) b) c) d)

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B/ absolute value expressions

 1. , a) b) c) 2. a) b) c)

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C/ one-sided limits

 1. a) b) c) 2. a) b)

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D/ piecewise defined functions

1. Let , find:

 a) b) c)

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2. Let , find:

 a) b) c)

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3. Let , find

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4. Let ,

a) find k so that f ( – 3) =

b) on the graph of f (x), what will the y-value be at the point where x = – 3?

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Solutions

A/

 1.a) ans: 22 b) ans: o c) = – 6 d) ans: 2.a) ans: 3/4 b) ans: 0 c) ans: – 3 3.a) factor, ans: 8 b) factor, ans: 12 c) factor, ans: 4 d) ans: 6

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B/ absolute value expressions (DNE means does not exist -- no limit)

 1. , a) b) c) DNE 2. a) b) c) DNE

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C/ one-sided limits

 1. a) b) c) 2. a) b)

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D/ piecewise defined functions

1. Let , find:

 a) b) c)

.

2. Let , find:

 a) b) c) DNE

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3. Let , find

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4. Let ,

a) find k so that f (–3) =

k should = – 6

b) on the graph of f (x), what will the y-value be at the point where x = – 3?

It will be – 6.

(all content of the MathRoom Lessons © Tammy the Tutor; 2004 - ).