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Finding limits algebraically - when direct substitution is not possible
- Lesson: 110:24
- Lesson: 23:55
- Lesson: 3a7:25
- Lesson: 49:43
- Lesson: 510:17
Finding limits algebraically - when direct substitution is not possible
There are times when applying direct substitution would only give us an undefined solution. In this section, we will explore some cool tricks to evaluate limits algebraically, such as using conjugates, trigonometry, common denominators, and factoring.
Lessons
- 1.Simplify Out "Zero Denominator" by Cancelling Common Factors
Find limx→3x−3x2−9
- 2.Expand First, Then Simplify Out "Zero Denominator" by Cancelling Common Factors
Evaluate limh→0h(5+h)2−25
- 3.Simplify Out "Zero Denominator" by Rationalizing Radicals
Evaluate:
a)limx→42−x4−x (hint: rationalize the denominator by multiplying its conjugate) - 4.Find Limits of Functions involving Absolute Value
Evaluate limx→0x∣x∣
(hint: express the absolute value function as a piece-wise function) - 5.Find Limits Using the Trigonometric Identity:limθ→0θsinθ=1
Find limx→02xsin5x
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16.
Limits
16.1
Finding limits from graphs
16.2
Continuity
16.3
Finding limits algebraically - direct substitution
16.4
Finding limits algebraically - when direct substitution is not possible
16.5
Infinite limits - vertical asymptotes
16.6
Limits at infinity - horizontal asymptotes
16.7
Intermediate value theorem
16.8
Squeeze theorem
Don't just watch, practice makes perfect
Practice topics for Limits
16.1
Finding limits from graphs
16.2
Continuity
16.3
Finding limits algebraically - direct substitution
16.4
Finding limits algebraically - when direct substitution is not possible
16.5
Infinite limits - vertical asymptotes
16.6
Limits at infinity - horizontal asymptotes
16.7
Intermediate value theorem
16.8
Squeeze theorem