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,
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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.
.
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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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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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 |
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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| algebraic limits | absolute value limits | one-sided limits | restricted domains |
| piecewise defined functions | practice | solutions |
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
| algebraic limits | absolute value limits | one-sided limits | restricted domains |
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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.
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However, if the function had been defined as:
instead,
the left limit , but the right limit .
The left limit is not equal to the right limit, so does not exist.
| algebraic limits | absolute value limits | one-sided limits | restricted domains |
| piecewise defined functions | practice | solutions |
Find the limits:
A/
1.a) | b) | c) | d) |
2.a) | b) | c) | |
3.a) | b) | c) | d) |
.
B/ absolute value expressions
1. , | a) | b) | c) |
2. | a) | b) | c) |
.
C/ one-sided limits
1. a) | b) | c) |
2. | a) | b) |
.
D/ piecewise defined functions
1. Let , find:
a) | b) | c) |
.
2. Let , find:
a) | b) | c) |
.
3. Let , find
.
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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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 |
.
B/ absolute value expressions (DNE means does not exist -- no limit)
1. , | a) | b) | c) DNE |
2. | a) | b) | c) DNE |
.
C/ one-sided limits
1. a) | b) | c) |
2. | a) | b) |
.
D/ piecewise defined functions
1. Let , find:
a) | b) | c) |
.
2. Let , find:
a) | b) | c) DNE |
.
3. Let , find
.
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.
| algebraic limits | absolute value limits | one-sided limits | restricted domains |
| piecewise defined functions | practice | solutions |
(all content of the MathRoom Lessons © Tammy the Tutor; 2004 - ).