Math Learning Center Supplements
698-1579
CB 116
EVALUATING LIMITS
An important application of determining the limit of a function is to understand the behavior of the graph
of that function. You can answer the questions, “What happens at very large values of x?”, “What happens at
very small values of x?”, and “Where are the horizontal or slant asymptotes?”
The idea of a limit might have been introduced first to you in a college algebra class when talking about
horizontal asymptotes. You may have been given the three cases for horizontal asymptotes:
•
𝑓(𝑥) =
CASE 1: If 𝑑𝑒𝑔𝑟𝑒𝑒 𝑎(𝑥) > 𝑑𝑒𝑔𝑟𝑒𝑒 𝑏(𝑥) then there is no horizontal asymptote
ex. 𝑓(𝑥) =
•
4𝑥 3 −4𝑥 2 +5
6𝑥 2 −7𝑥+2
The degree of the numerator 𝑎(𝑥) is 3 and the degree of the denominator 𝑏(𝑥) is 2.
There is no horizontal asymptote.
CASE 2: If 𝑑𝑒𝑔𝑟𝑒𝑒 𝑎(𝑥) < 𝑑𝑒𝑔𝑟𝑒𝑒 𝑏(𝑥) then the horizontal asymptote is 𝑦 = 0
ex. 𝑓(𝑥) =
•
𝑎(𝑥)
𝑏(𝑥)
2𝑥 3 −4𝑥 2 +5
6𝑥 5 −7𝑥 3 +2𝑥 2 −4𝑥+1
The degree of the numerator is 𝑎(𝑥) is 3 and the degree of the denominator 𝑏(𝑥) is 3.
The horizontal asymptote is 𝑦 = 0.
CASE 3: If 𝑑𝑒𝑔𝑟𝑒𝑒 𝑎(𝑥) = 𝑑𝑒𝑔𝑟𝑒𝑒 𝑏(𝑥) then the horizontal asymptote is the ratio of the coefficients.
ex. 𝑓(𝑥) =
4𝑥 3 −3𝑥 2 +2
7𝑥 3 +5𝑥 2 −𝑥+6
The degree of the numerator 𝑎(𝑥) is 3 and the degree of the denominator 𝑏(𝑥) is 3.
4
7
The horizontal asymptote is 𝑦 = , the ratio of the coefficients of the highest degree terms.
In each of these cases what is really being asked is “What does the graph approach at very small or very
large values of x?” Or in more technical terms:
What is lim𝑥→∞ 𝑓(𝑥) or lim𝑥→∞ 𝑓(𝑥)?
(the limit of the function as x approaches positive infinity and as x approaches negative infinity)
Math Learning Center Supplements
698-1579
CB 116
Instead of memorizing the three cases, we need only take the limit of the function as x approaches positive
or negative infinity. Lets look as the three cases again using limits.
•
CASE 1: If 𝑑𝑒𝑔𝑟𝑒𝑒 𝑎(𝑥) > 𝑑𝑒𝑔𝑟𝑒𝑒 𝑏(𝑥) then there is no horizontal asymptote
ex. 𝑓(𝑥) =
4𝑥 3 −4𝑥 2 +5
6𝑥 2 −7𝑥+2
This time we will take the limit as x approaches infinity.
lim𝑥→∞ 𝑓(𝑥) = lim𝑥→∞
4𝑥 3 −4𝑥 2 +5
6𝑥 2 −7𝑥+2
When you are evaluating infinite limits first determine the highest power of x in the function. In this case,
the highest power of x is to the third power. We can divide both the numerator and denominator by that
power.
lim𝑥→∞ 𝑓(𝑥) = lim𝑥→∞
= lim𝑥→∞
4𝑥 3 −4𝑥 2 +5
𝑥3
÷ 𝑥3
6𝑥 2 −7𝑥+2
4𝑥3 4𝑥2 5
− 3+ 3
𝑥3
𝑥
𝑥
6𝑥2 7𝑥 2
− 3+ 3
3
𝑥
𝑥
𝑥
4
5
4− + 3
𝑥 𝑥
=lim𝑥→∞ 6
𝟏
𝒙
7
2
− +
𝑥 𝑥2 𝑥3
We now must use the fact that as x gets very large gets closer and closer to 0.
=lim𝑥→∞
4−0−0
0−0+0
The limit is undefined. So there is no horizontal asymptote.
4
0
= = undefined / limit does not exist.
Math Learning Center Supplements
•
698-1579
CB 116
CASE 2: If 𝑑𝑒𝑔𝑟𝑒𝑒 𝑎(𝑥) < 𝑑𝑒𝑔𝑟𝑒𝑒 𝑏(𝑥) then the horizontal asymptote is 𝑦 = 0
ex. 𝑓(𝑥) =
2𝑥 3 −4𝑥 2 +5
6𝑥 5 −7𝑥 3 +2𝑥 2 −4𝑥+1
We can take the limit as x approaches infinity for this one as well.
2𝑥 3 −4𝑥 2 +5
lim𝑥→∞ 𝑓(𝑥) = lim𝑥→∞
6𝑥 5 −7𝑥 3 +2𝑥 2 −4𝑥+1
lim𝑥→∞ 𝑓(𝑥) = lim𝑥→∞
6𝑥 5 −7𝑥 3 +2𝑥 2 −4𝑥+1
This time the highest power of x is 5 so we must divide by 𝑥 5 .
= lim𝑥→∞
= lim𝑥→∞
2𝑥 3 −4𝑥 2 +5
2𝑥3 4𝑥2 5
− 5+ 5
𝑥
𝑥5
𝑥
6𝑥5 7𝑥3 2𝑥2 4𝑥 1
− 5 + 5 − 5+ 5
𝑥5
𝑥
𝑥
𝑥
𝑥
÷
𝑥5
𝑥5
2
4
5
− +
𝑥2 𝑥3 𝑥5
7
2
4
1
6− 2+ 3 − 4 + 5
𝑥
𝑥
𝑥
𝑥
0−0+0
0
= 6−0+0−0+0 = 6 = 0
The limit is 0 which means our graph approaches 𝑦 = 0 for extreme values of x. So 𝑦 = 0 is the
horizontal asymptote.
Math Learning Center Supplements
•
698-1579
CB 116
CASE 3: If 𝑑𝑒𝑔𝑟𝑒𝑒 𝑎(𝑥) = 𝑑𝑒𝑔𝑟𝑒𝑒 𝑏(𝑥) then the horizontal asymptote is the ratio of the coefficients.
ex. 𝑓(𝑥) =
4𝑥 3 −3𝑥 2 +2
7𝑥 3 +5𝑥 2 −𝑥+6
Let’s take the limit as x approaches negative infinity.
lim𝑥→−∞ 𝑓(𝑥) = lim𝑥→−∞
4𝑥3 −3𝑥2 +2
7𝑥3 +5𝑥2 −𝑥+6
The highest power of 𝑥 is to the third power so we will divide by 𝑥 3 .
4𝑥 3 −3𝑥 2 +2
𝑥3
lim𝑥→−∞ 𝑓(𝑥) = lim𝑥→−∞ 7𝑥 3 +5𝑥 2 −𝑥+6 ÷ 𝑥 3
= lim𝑥→−∞
4𝑥3 3𝑥2 2
− 3+ 3
𝑥3
𝑥
𝑥
7𝑥3 5𝑥2 𝑥
6
+
−
+
𝑥3
𝑥3 𝑥3 𝑥3
= lim𝑥→−∞
7+ − 2 + 3
𝑥 𝑥
𝑥
1
3
2
4− + 3
𝑥 𝑥
5
1
6
As 𝑥 approaches negative infinity 𝑥 approaches 0. It approaches 0 from below but it still approaches 0.
4−0+0
4
= 7+0−0+0 = 7
4
The horizontal asymptote for this function is 𝑦 = 7.
Math Learning Center Supplements
698-1579
CB 116
We have confirmed the three cases. Now let’s try evaluating other limits.
FIRST:
When solving limits, first try simply to substitute the number into the function and evaluate.
Sometimes this is all you need to do.
1.
𝑥−4
lim𝑥→2 𝑥+6
𝑥−4
2.
3.
2−4
2
1
1
lim𝑥→2 𝑥+6 = 2+6 = − 8 = − 4 so the answer is − 4.
lim𝑥→6 𝑥 2 + 3𝑥 − 5
lim𝑥→6 𝑥 2 + 3𝑥 − 5 = (6)2 + 3(6) − 5 = 36 + 18 − 5 = 49 so the answer is 49.
lim𝑥→−∞ −𝑥 3 + 5
lim𝑥→−∞ −𝑥 3 + 5 = −(−∞)3 + 5 = −(−∞) + 5 = ∞ + 5 = ∞, the limit does not exist.
SECOND:
Sometimes it is not possible to just evaluate the function because the function is not defined for that
value. If this happens then try to simplify (factor or divide) the function before evaluating.
4.
lim𝑥→1
𝑥 2 +4𝑥−5
= lim𝑥→1
5.
𝑥−1
(𝑥−1)(𝑥+5)
𝑥−1
= lim𝑥→1
(𝑥−1)(𝑥+5)
𝑥−1
= lim𝑥→1 𝑥 + 5
= lim𝑥→1 𝑥 + 5 = 1 + 5 = 6 so the answer is 6.
lim𝜃→0
tan2 𝜃
𝜃
sin2 𝜃
lim𝜃→0 𝜃 cos2 𝜃 = lim𝜃→0
lim𝜃→0
sin 𝜃
𝜃
sin 𝜃
sin 𝜃
𝜃
sin 𝜃
∙ cos2 𝜃
0
∙ lim𝜃→0 cos2 𝜃 = 1 ∙ 1 = 0
THIRD:
Whenever you are dealing with positive or negative infinity and a rational function (fractional), if you
are unable to factor, then divide each piece by the highest power of x in the function.
6.
2𝑥 3 −3𝑥+6
lim𝑥→∞ 3𝑥 5 +4𝑥 3 −6𝑥−1
2𝑥 3 −3𝑥+6
lim𝑥→∞ 3𝑥 5 +4𝑥 3 −6𝑥−1 = lim𝑥→∞
=0
2
3
6
− +
𝑥2 𝑥4 𝑥5
4
6
1
3+ 2 − 4 − 5
𝑥
𝑥
𝑥
0
=3
Math Learning Center Supplements
698-1579
CB 116
FOURTH:
When all else fails, graph the function.
7.
lim𝑥→3
|𝑥−3|
𝑥−3
In this function the absolute value sign may throw us off. Is the limit 1? Graphing the function
makes the answer more clear.
The limit from the left is -1.
The limit from the right is +1.
lim𝑥→3−
lim𝑥→3+
Since the two limits are not equal lim𝑥→3
−1
8. lim𝑥→−1 𝑥+1
The limit from the left is −∞.
The limit from the right is +∞.
|𝑥−3|
𝑥−3
|𝑥−3|
|𝑥−3|
𝑥−3
𝑥−3
= −1.
= 1.
does not exist.
−2𝑥
lim𝑥→−1− 𝑥+1 = −∞
−2𝑥
lim𝑥→−1+ 𝑥+1 = +∞
−2𝑥
Since the two limits are not equal lim𝑥→−1 𝑥+1 does not exist.
−1
9. lim𝑥→2 (𝑥−2)2
The limit from the left is −∞.
The limit from the right is −∞.
−1
−1
lim𝑥→2− (𝑥−2)2 = −∞.
−1
lim𝑥→2+ (𝑥−2)2 = −∞.
Thus lim𝑥→2 (𝑥−2)2 = −∞, the limit does not exist.
10. lim𝜃→0
1−cos 𝜃
𝜃
The limit from the left is 0.
The limit from the right is 0.
Thus lim𝜃→0
1−cos 𝜃
𝜃
= 0.
lim𝜃→0−
lim𝜃→0+
1−cos 𝜃
𝜃
1−cos 𝜃
𝜃
= 0.
= 0.