Table of contents

0. Functions7h 52m
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 Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

1. Limits and Continuity

Continuity

Calculus

1. Limits and Continuity

Continuity

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2:40 minutes

Problem 2.5.42

Textbook Question

Define h(2) in a way that extends h(t) = (t² + 3t − 10)/(t − 2) to be continuous at t = 2.

Verified step by step guidance

First, identify the function h(t) = (t² + 3t − 10)/(t − 2). Notice that the denominator becomes zero when t = 2, which makes the function undefined at this point.

To extend h(t) to be continuous at t = 2, we need to find the limit of h(t) as t approaches 2. This involves simplifying the expression to remove the discontinuity.

Factor the numerator t² + 3t − 10. Look for two numbers that multiply to -10 and add to 3. These numbers are 5 and -2, so the factorization is (t + 5)(t - 2).

Substitute the factorized form into the function: h(t) = ((t + 5)(t - 2))/(t - 2). Notice that the (t - 2) terms cancel out, simplifying the function to h(t) = t + 5 for t ≠ 2.

Now, find the limit of h(t) as t approaches 2 using the simplified function: lim(t→2) (t + 5). This limit will give the value of h(2) that makes the function continuous at t = 2.

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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. To define h(2) for continuity, we need to evaluate the limit of h(t) as t approaches 2. If this limit exists, it can be used to assign a value to h(2) that makes the function continuous at that point.

Continuity

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. For h(t) to be continuous at t = 2, we must ensure that h(2) is defined and equals the limit of h(t) as t approaches 2. This ensures there are no breaks or jumps in the function at that point.

Rational Functions

Rational functions are ratios of polynomials, and they can have points of discontinuity where the denominator equals zero. In this case, h(t) has a denominator of (t - 2), which becomes zero at t = 2, indicating a potential discontinuity. To extend h(t) to be continuous at t = 2, we need to simplify the function and find a suitable value for h(2) that resolves this discontinuity.