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
- 8. Definite Integrals3h 25m
1. Limits and Continuity
Finding Limits Algebraically
Problem 27a
Textbook Question
Determine the following limits.
a. lim x→2^+ x^2 − 4x + 3 / (x − 2)^2

1
Step 1: Identify the type of limit problem. This is a one-sided limit as \( x \) approaches 2 from the right (\( x \to 2^+ \)).
Step 2: Substitute \( x = 2 \) into the function \( \frac{x^2 - 4x + 3}{(x - 2)^2} \) to check if it results in an indeterminate form. Substituting gives \( \frac{2^2 - 4 \times 2 + 3}{(2 - 2)^2} = \frac{4 - 8 + 3}{0} = \frac{-1}{0} \), indicating a division by zero.
Step 3: Analyze the behavior of the numerator and denominator as \( x \to 2^+ \). The numerator \( x^2 - 4x + 3 \) simplifies to \( (x - 1)(x - 3) \). As \( x \to 2^+ \), \( x - 1 \to 1 \) and \( x - 3 \to -1 \), so the numerator approaches \( 1 \times -1 = -1 \).
Step 4: Consider the denominator \( (x - 2)^2 \). As \( x \to 2^+ \), \( x - 2 \to 0^+ \), so \( (x - 2)^2 \to 0^+ \).
Step 5: Determine the limit by combining the behavior of the numerator and denominator. Since the numerator approaches \(-1\) and the denominator approaches \(0^+\), the limit approaches \(-\infty\).

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Limits
Limits are fundamental in calculus, representing the value that a function approaches as the input approaches a certain point. They help in understanding the behavior of functions near specific points, including points of discontinuity or indeterminate forms. Evaluating limits is crucial for defining derivatives and integrals.
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Indeterminate Forms
Indeterminate forms occur when direct substitution in a limit leads to expressions like 0/0 or ∞/∞, which do not provide clear information about the limit's value. In such cases, techniques like factoring, rationalizing, or applying L'Hôpital's Rule are used to resolve these forms and find the limit.
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Continuous Functions
A function is continuous at a point if the limit of the function as it approaches that point equals the function's value at that point. Understanding continuity is essential when evaluating limits, as it allows for the direct substitution of values in many cases, simplifying the limit evaluation process.
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