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Chapter 1: Q. 34 (page 135)

Calculate each of the limits in Exercises $29 - 70$ .
$\lim_{x \to - 2} \frac{4 + 2 x}{x + 2}$ .

Short Answer

Expert verified

The value of $\lim_{x \to - 2} \frac{4 + 2 x}{x + 2}$ is, $2$ .

Step by step solution

01

Step 1. Given information

$\lim_{x \to - 2} \frac{4 + 2 x}{x + 2}$ .

02

Step 2. The limit is calculated as below:

$\lim_{x \to - 2} \frac{4 + 2 x}{x + 2} = \lim_{x \to - 2} \frac{2 (x + 2)}{(x + 2)} = \lim_{x \to - 2} 2 = 2$

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Most popular questions from this chapter

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 $\lim_{x \to c} f (x) = L$ in Exercises $47 - 60$ .

$\lim_{x \to 3} (x^{2} - 6 x + 11) = 2$ .

Write delta-epsilon proofs for each of the limit statements $\lim_{x \to c} f (x) = L$ in Exercises $47 - 60$ .

$\lim_{x \to 2} (3 - 4 x) = - 5$ .

Write each of the inequalities in interval notation:

$|(x^{2} - 1) + 3| < 0.5$

Calculate each of limits:

$\lim_{x \to 1} \frac{1}{\sin^{- 1} x}$ .

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