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Chapter 1: Q. 64 (page 136)

Calculate each of the limits:
$\lim_{h \to 0} \frac{{(2 + h)}^{2} - 2^{2}}{h}$ .

Short Answer

Expert verified

The solution of the limit $\lim_{h \to 0} \frac{{(2 + h)}^{2} - 2^{2}}{h}$ is, $4$ .

Step by step solution

01

Step 1. Given Information.

The limit is,

$\lim_{h \to 0} \frac{{(2 + h)}^{2} - 2^{2}}{h}$ .

02

Step 2. Finding the solution.

$\lim_{h \to 0} \frac{{(2 + h)}^{2} - 2^{2}}{h}$

Using ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ ,

$= \lim_{h \to 0} \frac{h^{2} + 4 h + 4 - 4}{h} = \lim_{h \to 0} \frac{h^{2} + 4 h}{h} = \lim_{h \to 0} h + 4$

Inputting $h = 0$

$= 0 + 4 = 4$

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Most popular questions from this chapter

What are punctured intervals, and why do we need to use them when discussing limits?

Each function in Exercises 9–12 is discontinuous at some value x = c. Describe the type of discontinuity and any one-sided continuity at x = c, and sketch a possible graph of f.

$\lim_{x \to 0^{-}} f (x) = - 1, \lim_{x \to 0^{+}} f (x) = 1, f (0) = 0 .$

Write a delta–epsilon proof that proves that $f$ is continuous on its domain. In each case, you will need to assume that δ is less than or equal to $1$ .

$f (x) = x^{2}$

Write delta-epsilon proofs for each of the limit statements in Exercises $47 - 60 .$

$\lim_{x \to 5^{+}} \sqrt{x - 5} = 0$

Find a formula for the cost $C (r)$ of producing a gourmet soup can with radius $r$ and height $5$ inches, and answer the following questions:

What is the radius of a can that is $5$ inches tall and costs $30$ cents to produce?
Your manager wants you to produce $5$ -inch-tall cans that cost between $20$ and $40$ cents. Write this requirement as an absolute value inequality.
What range of radii would satisfy your manager? Write an absolute value inequality whose solution set lies inside this range of radii.

See all solutions

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