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Chapter 1: Q. 5 (page 153)

Calculating limits: Find each limit by hand.
$\lim_{x 鈫� 1} \frac{2 x^{3} - x^{2} - 2 x + 1}{x^{2} - 2 x + 1}$ .

Short Answer

Expert verified

The answer of the limit is $鈭�$ .

Step by step solution

01

Step 1. Given Information.

Given: $\lim_{x 鈫� 1} \frac{2 x^{3} - x^{2} - 2 x + 1}{x^{2} - 2 x + 1}$ .

02

Step 2. Calculating the limit using direct substitution.

$S o l v i n g b y h a n d, \lim_{x 鈫� 1} \frac{2 x^{3} - x^{2} - 2 x + 1}{x^{2} - 2 x + 1} = \lim_{x 鈫� 1} \frac{(x - 1) (x + 1) (2 x - 1)}{{(x - 1)}^{2}} = \lim_{x 鈫� 1} \frac{(x + 1) (2 x - 1)}{(x - 1)} = \frac{2 脳 1}{0} = 鈭� .$

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Most popular questions from this chapter

For each limit statement , use algebra to find 未 > 0 in terms of $蔚$ > 0 so that if 0 < |x 鈭� c| < 未, then | f(x) 鈭� L| < $蔚$ .

$\lim_{x 鈫� 2} (x^{2} - 4 x + 6) = 2$

Use the delta-epsilon definition of continuity to argue that f is or is not continuous at the indicated point $x = c$ .

$f (x) = \{\begin{aligned} 2 鈭� x, 鈥呪赌呪赌呪赌� if x rational \\ x^{2}, 鈥呪赌呪赌呪赌� if x irrational, \end{aligned} c = 2$

For each limit statement , use algebra to find 未 > 0 in terms of $蔚$ > 0 so that if 0 < |x 鈭� c| < 未, then | f(x) 鈭� L| < $蔚$ .

$\lim_{x 鈫� 3} \frac{1}{x} = \frac{1}{3}; you may assume 未鈮� 1$

For each limit in Exercises 33鈥�38, either use continuity to calculate the limit or explain why Theorem 1.16 does not apply.

$\lim_{x 鈫� - 5} \sqrt{x} .$

$\lim_{x 鈫� 1} \frac{1 - \cos (x - 1)}{x}$

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