Ch. 2 - Limits and Continuity

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 2 - Limits and Continuity

Problem 2.5.42

Chapter 2, Problem 2.5.42

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.

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Related Practice

Textbook Question

Finding Limits

In Exercises 3–8, find the limit of each function (a) as x → ∞ and (b) as x → −∞. (You may wish to visualize your answer with a graphing calculator or computer.)

h(x) = (−5 + (7/x))/(3 – (1/x²))

Textbook Question

Graphing Simple Rational Functions

Graph the rational functions in Exercises 63–68. Include the graphs and equations of the asymptotes and dominant terms.

y = 1/(x − 1)

Textbook Question

Using the Formal Definitions

Use the formal definitions of limits as x → ±∞ to establish the limits in Exercises 91 and 92.

If f has the constant value f(x) = k, then lim x → ∞ f(x) = k.

Textbook Question

Using Limit Rules

Suppose lim x→0 f(x) = 1 and lim x→0 g(x) = −5. Name the rules in Theorem 1 that are used to accomplish steps (a), (b), and (c) of the following calculation.

limx→0 (2f(x) − g(x)) / (f(x) + 7)² = limx→0 (2f(x) − g(x)) / limx→0 (f(x) + 7)² (a)

(We assume the denominator is nonzero.)

(lim x→0 2f(x) − lim x→0 g(x)) / (lim x→0 (f(x) + 7))² (b)

= (2 lim x→0 f(x) − lim x→0 g(x)) / (lim x→0 f(x) + lim x→0 7)² (c)

= ((2)(1) − (−5)) / (1 + 7)² = 7/64

Textbook Question

For what values of a and b is

g(x) = { ax + 2b, x ≤ 0

x² + 3a – b, 0 < x ≤ 2

3x – 5, x > 2

continuous at every x?

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Textbook Question

Limits as x → ∞ or x → −∞

The process by which we determine limits of rational functions applies equally well to ratios containing noninteger or negative powers of x. Divide numerator and denominator by the highest power of x in the denominator and proceed from there. Find the limits in Exercises 23–36. Write ∞ or −∞ where appropriate.

lim x → ⁻∞ ((1 − x³) / (x² + 7x))⁵