Evaluating Limits Analytically

EVALUATING LIMITS ANALYTICALLY
Last class, we learned that the limit of 𝑓(π‘₯) as π‘₯ approaches π‘Ž does not depend on
the value of 𝑓 at π‘₯ = π‘Ž.
It may happen, however, that the limit is precisely 𝑓 π‘Ž , which means
lim 𝑓 π‘₯ = 𝑓(π‘Ž)
π‘₯β†’π‘Ž
In such cases, the limit can be evaluated by direct substitution.
THEOREM 2.1 Some Basic Limits
Let π‘Ž and 𝑏 be real numbers and let 𝑛 be a positive integer. Then,
1.
lim 𝑏 = 𝑏
π‘₯β†’π‘Ž
2.
lim π‘₯ = π‘Ž
π‘₯β†’π‘Ž
3.
lim π‘₯ 𝑛 = π‘Žπ‘›
π‘₯β†’π‘Ž
Properties of Limits Handout
Ex 3. Let:
lim 𝑓 π‘₯ = 4
π‘₯β†’π‘Ž
Find:
lim [𝑓 π‘₯ + β„Ž π‘₯ ]
a) π‘₯β†’π‘Ž
𝑓 π‘₯
b) lim
π‘₯β†’π‘Ž β„Ž π‘₯
lim 𝑔 π‘₯ = 0
lim β„Ž π‘₯ = βˆ’3
π‘₯β†’π‘Ž
π‘₯β†’π‘Ž
c)
d)
lim β„Ž π‘₯
2
π‘₯β†’π‘Ž
β„Ž π‘₯
lim
π‘₯β†’π‘Ž 𝑔 π‘₯
So, if it’s a polynomial or rational with
nonzero denominator, just plug it in…
In fact, always try to plug in π‘Ž and see what happens…
If you get a number back, then great! That’s the limit!
If not…
DIVIDING OUT (FACTORING) TECHNIQUE
π‘₯2 + π‘₯ βˆ’ 6
Ex 4. lim
π‘₯β†’βˆ’3
π‘₯+3
COMMON DENOMINATOR TECHNIQUE
Ex 5.
1
1
βˆ’
lim π‘₯ + 4 4
π‘₯β†’0
π‘₯
RATIONALIZING (CONJUGATE) TECHNIQUE
Ex 6.
π‘₯+1βˆ’1
lim
π‘₯β†’0
π‘₯
THEOREM 2.9 Two Special Limits
1.
sin π‘₯
lim
=1
π‘₯β†’0 π‘₯
2.
1 βˆ’ cos π‘₯
lim
=0
π‘₯β†’0
π‘₯
Ex 7.
tan π‘₯
lim
π‘₯β†’0 π‘₯
Ex 8.
sin 4π‘₯
lim
π‘₯β†’0 5π‘₯
Ex 9.
2βˆ’π‘₯
lim+
π‘₯β†’2 |2 βˆ’ π‘₯|
So, in summary:
1.
2.
3.
4.
5.
Try Direct Substitution! You might get lucky! If not…
Factoring (Dividing Out)
Common Denominator (if you see fractions in fractions)
Rationalizing (if you see a radical)
Remember special trig limits!