Download evaluating limits

East Campus, CB 117
361-698-1579
Math Learning Center
EVALUATING LIMITS
West Campus, HS1 203
361-698-1860
STEPS TO EVALUATING LIMITS:
I.
Always start by plugging in the number
A. If you get an answer, then you can stop
0 ?
∞
B. If you get 0 , 0, or ∞ then you need to do something else.
0
1. If you get 0, then
a. Factor the numerator and denominator and cancel out like factors. Then try to plug the
number in. If you get an answer, then that’s your answer and you can stop.
b. If you cannot factor, then test from each side.
?
2. If you get 0, then test from each side.
∞
3. If you get ∞, then divide the numerator and denominator by the variable with the largest
exponent in the denominator. Then simplify and then plug the number in.
EXAMPLES:
Example 1: lim 5𝑥𝑥 + 6 using step I., try plugging in the number → 5(3) + 6 → 15 + 6 → 𝟐𝟐𝟐𝟐, so 21 is the
𝑥𝑥→3
answer and you’re done.
Example 2: lim 2 sin 𝑥𝑥 + cos 𝑥𝑥 using step I., try plugging in the number
𝑥𝑥→𝜋𝜋
2 sin(𝜋𝜋) + cos(𝜋𝜋) → 2(0) + −1 → 0 + −1 → −𝟏𝟏
Example 3: lim
𝑥𝑥 2−16
𝑥𝑥→4 𝑥𝑥−4
→
lim
→
42 −16
16−16
0
→ 4−4 → 0 →by step I.B.1. factor
(4)−4
in the number.
𝑥𝑥 2−16
𝑥𝑥→4 𝑥𝑥−4
Example 4: lim
|𝑦𝑦+2|
𝑦𝑦→−2 𝑦𝑦+2
lim
|𝑦𝑦+2|
𝑦𝑦→−2 𝑦𝑦+2
lim −
𝑦𝑦→ −2
(𝑥𝑥−4)(𝑥𝑥+4)
(𝑥𝑥−4)
→
|−2+2|
−2+2
→
(𝑥𝑥−4)(𝑥𝑥+4)
(𝑥𝑥−4)
0
the top and bottom and then try plugging
→ (𝑥𝑥 + 4) → 4 + 4 → 𝟖𝟖
→ 0 → since you cannot factor, by step I.B.1.b. test from each side.
→ use a number really close to -2 from the left side, like -2.001 for instance. Use this
|𝑦𝑦+2|
𝑦𝑦+2
number to plug into the fraction to see where it is approaching.
|−2.001 + 2|
−2.001+2
→
|−.001|
−1.001
→ use a number really close to -2 from the left side, like -1.999 for instance. Use
this number to plug into the fraction to see where it is approaching.
|−1.999+2|
−1.999+2
→
|.001|
.001
Since lim − ≠ lim + then lim
𝑦𝑦→−2
.001
→ −.001 → −1
𝑦𝑦→−2
.001
→ .001 → 1
|𝑦𝑦+2|
𝑦𝑦→−2 𝑦𝑦+2
Does Not Exist
East Campus, CB 117
361-698-1579
4
Math Learning Center
4
4
Example 5: lim 𝑥𝑥−7 → 7−7 → 0 → by step I.B.2. test from each side.
𝑥𝑥→7
lim
4
𝑥𝑥→7− 𝑥𝑥−7
4
lim
𝑥𝑥→7+ 𝑥𝑥−7
→ use a number really close to 7 from the left side, like 6.999 for instance. Use
this number to plug into the fraction to see where it is approaching.
4
4
→ −.001 → −4000. Therefore, as x gets even closer to 7, the limit is going to
6.999−7
approach −∞.
→ use a number really close to 7 from the right side, like 7.001 for instance. Use this
number to plug into the fraction to see where it is approaching.
4
4
→ .001 → 4000. Therefore, as x gets even closer to 7, the limit is going to
7.001−7
approach +∞.
4
Does Not Exist.
Since lim− ≠ lim+ then lim 𝑥𝑥−7
𝑦𝑦→7
Example 6: lim
West Campus, HS1 203
361-698-1860
𝑦𝑦→7
𝑥𝑥 2+3
𝑦𝑦→7
(∞)2 +3
∞
→ 3(∞)2 −(∞)+2 → ∞ → by step I.B.3. divide the numerator and denominator by the
𝑥𝑥→∞
3𝑥𝑥 2−𝑥𝑥+2
lim
𝑥𝑥2 𝑥𝑥2
+
𝑥𝑥2 𝑥𝑥2
2
3𝑥𝑥
𝑥𝑥
2
− +
𝑥𝑥2 𝑥𝑥2 𝑥𝑥2
variable with the largest exponent that occurs in the denominator, which in this case is 𝑥𝑥 2 , and
then plug in the number
𝑥𝑥→∞
→
3
1+ 2
𝑥𝑥
1
2
3− + 2
𝑥𝑥 𝑥𝑥
→
3
1+ 2
∞
1
→
2
3− + 2
∞ ∞
3
∞
1+
1 2
∞ ∞
3− +
Since any constant divided by a really really large number is approaching zero � lim
then
3
∞
1+
1 2
3− +
∞ ∞
1+0
𝑥𝑥 3+3
Example 7: lim
1
→ 3−0+0 → 3 thus lim
𝑥𝑥→∞ 3𝑥𝑥 2−𝑥𝑥+2
𝑥𝑥→∞
(∞)3 +3
→ lim
𝑥𝑥 2+3
3𝑥𝑥 2−𝑥𝑥+2
𝑥𝑥→∞ 3(∞)2−(∞)+2
→
∞
→
1
𝑥𝑥→∞ 𝑥𝑥
𝟏𝟏
= 0�,
𝟑𝟑
∞
→ by step I.B.3. divide the numerator and denominator
→
3
∞
1 2
3− +
∞ ∞
by the variable with the largest exponent that occurs in the denominator, which in this case is
𝑥𝑥 2 , and then plug in the number
lim
𝑥𝑥→∞
𝑥𝑥3 3
+
𝑥𝑥2 𝑥𝑥2
3𝑥𝑥2 𝑥𝑥
2
− +
𝑥𝑥2 𝑥𝑥2 𝑥𝑥2
→
3
𝑥𝑥+ 2
𝑥𝑥
1 2
3− + 2
𝑥𝑥 𝑥𝑥
→
3
∞+ 2
∞
1
2
3− + 2
∞ ∞
∞+
Since any constant divided by a really really large number is approaching zero � lim
then
3
∞
1 2
3− +
∞ ∞
∞+
∞+0
→ 3−0+0 →
∞
3
→ ∞ thus lim
1
𝑥𝑥→∞ 𝑥𝑥
𝑥𝑥 3+3
𝑥𝑥→∞ 3𝑥𝑥 2−𝑥𝑥+2
→∞
= 0�,