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Chapter 1: Q. 19 (page 153)

Consider the limit expression
$\lim_{x 鈫� 鈭�} \frac{(2 x - 1) (x^{2} + 1)}{x^{2} - 4}$
Calculate the limit.

Short Answer

Expert verified

The limit of the expression is $鈭�$ .

Step by step solution

01

Step 1. Given

The given expression is

$\lim_{x 鈫� 鈭�} \frac{(2 x - 1) (x^{2} + 1)}{x^{2} - 4}$

02

Step 2. Calculation

The evaluated expression is

$\lim_{x 鈫� 鈭�} \frac{(2 x - 1) (x^{2} + 1)}{x^{2} - 4} = \lim_{x 鈫� 鈭�} \frac{x (2 - \frac{1}{x}) x (x + \frac{1}{x})}{x^{2} (1 - \frac{4}{x^{2}})} = \lim_{x 鈫� 鈭�} \frac{(2 - \frac{1}{x}) (x + \frac{1}{x})}{(1 - \frac{4}{x^{2}})} = \frac{(2 - \frac{1}{鈭�}) (鈭� + \frac{1}{鈭�})}{(1 - \frac{4}{{鈭�}^{2}})} = \frac{(2 - 0) (鈭� + 0)}{(1 - 0)} = \frac{2 鈭�}{1} = 鈭�$

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Most popular questions from this chapter

$\lim_{x 鈫� 1} (\frac{e^{x} - 1}{e^{2 x} + 2 e^{x} - 3})$

Calculate each of the limits:

$\lim_{x 鈫� 3} \frac{\sqrt{x}}{\tan^{- 1} \sqrt{x}}$

Write each of the inequalities in interval notation:

$0 < | x + 3 | < 0.05$

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

$f (x) = x^{鈭� 1 / 2}$

Each function in Exercises 9鈥�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.

$\lim_{x 鈫� - 1^{-}} f (x) = 2, \lim_{x 鈫� - 1^{+}} f (x) = 2, f (- 1) = 1 .$

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