Question 2: How do you evaluate limits at infinity using algebra?
Limits involving rational expression may be evaluated algebraically. To do this, we need
to make an observation.
For any positive real number n,
1
1
 0 and lim n  0
n
x  x
x  x
lim
as long as xn is defined.
Consider the case where n  2 . A table of values for x2 and
1
indicates how the parts
x2
of the rational expression behave.
x
10
100
1000
x2
100
10000
1000000
1
x2
0.01
0.0001
0.0000001
As x grows larger in each limit, the denominator x2 also grows larger. This means the
fraction
1
grows smaller and smaller. When evaluating rational expressions, our goal
x2
is to simplify the terms in the expression so we can see which terms become smaller
and smaller.
We need to be careful about these limits since the second limit is undefined for some
values of n. For some values of n like n  12 , xn is undefined for negative x values since
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this amounts to taking an even root of a negative number. In these cases, the second
limit is not defined.
Example 4
Evaluate the Limit
Evaluate the limit algebraically: lim
x 
2x 1
5x  4
Solution Start by finding the highest power that appears on x in the
denominator. The highest power on the variable in 5 x  4 is one. Divide
each term in the fraction by x to this power to yield
2x 1

2x 1
 lim x x
lim
x  5 x  4
x  5 x
4

x x
1
2
x
 lim
x 
4
5
x
For larger and larger values of x, the fractions
1
4
and
get smaller and
x
x
smaller. As these terms approach zero, the constant terms are
unchanged. The value of the limit is
0
1
2
2x 1
x2
 lim
lim
x  5 x  4
x 
4 5
5
x
0
The arrows help us to see how the individual pieces drop out as x gets
larger.
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Example 5
Evaluate the Limit
Evaluate the limit algebraically: lim
x 
2x
x  3x  2
2
Solution The highest power in the denominator is two. Divide each term
in the rational expression by x2 and examine the resulting terms:
2x
2
2x
lim 2
 lim 2 x
x  x  3 x  2
x  x
3x 2
 2 2
2
x
x
x
0
2
x
 lim
x 
3 2
1  2
x x
0
0
0
Each of the terms in red get small as x increases. This means the
denominator approaches 1 but the numerator approaches 0.
Example 6
Evaluate the Limit
10 x3  x  1
x 
2 x2  1
Evaluate the limit algebraically: lim
Solution The highest power of the variable that appears in the
denominator is two. Divide each term in the rational expression to give
10 x3 x 1
 2 2
2
10 x3  x  1
x
x
x
lim
 lim
x 
x 
2 x2 1
2x2 1

x2 x2
0
 lim
x 
1 1

x x2
1
2 2
x
0
10 x 
0
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Each of the terms in red grow smaller and smaller. However, the term in
blue grows more and more negative as x grows more and more
negative. If the numerator grows more and more negative, the fraction
becomes more and more negative. The limit does not exist. Since it
does this by becoming more and more negative, we write
10 x3  x  1
 
lim
x 
2x2 1
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