Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
1. Limits and Continuity
Finding Limits Algebraically
Problem 70b
Textbook Question
Find the vertical asymptotes. For each vertical asymptote x=a, analyze lim x→a- f(x) and lim x→a+ f(x).
f(x) = (x2 − 4x + 3) / (x − 1)
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1
Step 1: Identify the points where the denominator is zero, as these are potential vertical asymptotes. Set the denominator equal to zero: x - 1 = 0, which gives x = 1.
Step 2: Confirm that x = 1 is a vertical asymptote by checking that the numerator does not also become zero at x = 1. Substitute x = 1 into the numerator: (1)^2 - 4(1) + 3 = 0, which means the numerator is zero, so x = 1 is not a vertical asymptote but a removable discontinuity.
Step 3: Factor the numerator to simplify the expression. The numerator x^2 - 4x + 3 can be factored as (x - 1)(x - 3).
Step 4: Simplify the function by canceling the common factor (x - 1) from the numerator and the denominator, resulting in f(x) = x - 3 for x ≠ 1.
Step 5: Since the factor (x - 1) cancels out, there are no vertical asymptotes. However, analyze the behavior around x = 1 to understand the nature of the discontinuity.
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