ALGEBRAIC LIMITS 
A LIMIT IS A YVALUE 
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 yvalue 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 .
.
 algebraic limits  absolute value limits  onesided limits  restricted domains 
 piecewise defined functions  practice  solutions 
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 onesided limits.
So, for , we define the leftside and rightside limits.(see diagram)
..... and .....
Since .... , we say that does not exist.
.
 algebraic limits  absolute value limits  onesided limits  restricted domains 
 piecewise defined functions  practice  solutions 
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 leftside 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 rightside 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

.
 algebraic limits  absolute value limits  onesided 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 onesided limits for the extremes of this function. and
 algebraic limits  absolute value limits  onesided limits  restricted domains 
 piecewise defined functions  practice  solutions 
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.
.
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  onesided 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/ onesided 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 yvalue be at the point where x = – 3?
.
 algebraic limits  absolute value limits  onesided limits  restricted domains 
 piecewise defined functions  practice  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 
.
B/ absolute value expressions (DNE means does not exist  no limit)
1. ,  a)  b)  c) DNE 
2.  a)  b)  c) DNE 
.
C/ onesided 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 yvalue be at the point where x = – 3?
It will be – 6.
 algebraic limits  absolute value limits  onesided limits  restricted domains 
 piecewise defined functions  practice  solutions 
(all content of the MathRoom Lessons © Tammy the Tutor; 2004  ).