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Chapter 1: Q. 61 (page 149)

Calculate each limit in Exercises 35鈥�80.
$\underset{x 鈫� 鈭�}{l i m} \frac{2 e^{1.5 x}}{3 e^{2 x} + e^{1.5 x}}$

Short Answer

Expert verified

The limit is 0

Step by step solution

01

Step 1. Given information

The given expression is $\underset{x 鈫� 鈭�}{l i m} \frac{2 e^{1.5 x}}{3 e^{2 x} + e^{1.5 x}}$

02

Step 2. Calculation

The limit is calculated as below,

$\underset{x 鈫� 鈭�}{l i m} \frac{2 e^{1.5 x}}{3 e^{2 x} + e^{1.5 x}} = \underset{x 鈫� 鈭�}{l i m} \frac{2 e^{1.5 x}}{3 e^{2 x} + e^{1.5 x}} 脳 \frac{\frac{1}{e^{1.5 x}}}{\frac{1}{e^{1.5 x}}} = \underset{x 鈫� 鈭�}{l i m} \frac{2}{3 e^{2 x - 1.5 x} + 1} = \underset{x 鈫� 鈭�}{l i m} \frac{2}{3 e^{0.5 x} + 1} = \frac{2}{3 e^{0.5 (鈭�)} + 1} = \frac{2}{3 (鈭�) + 1} = 0$

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Most popular questions from this chapter

For each limit $\underset{x 鈫� c}{l i m} f (x) = L$ in Exercises 43鈥�54, use graphs and algebra to approximate the largest value of $未$ such that if $x 鈭� (c - 未, c) 鈭� (c, c + 未) then f (x) 鈭� (L - 蔚, L + 蔚)$

$\underset{x 鈫� - 1}{l i m} \frac{x^{2} - 2 x - 3}{x + 1} = - 4, 蔚 = 0.1$

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| < $蔚$ .

$\lim_{x 鈫� 3} (x^{2} - 2 x - 3) = 0; you may assume 未鈮� 1$

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.

$f (x) = x^{鈭� 1}$

Use algebra to find the largest possible value of 未 or smallest possible value of N that makes each implication true. Then verify and support your answers with labeled graphs.

$If x > N, then |\frac{1}{x^{2}} 鈭� 0| < 0.001$

Sketch a labeled graph of a function that satisfies the hypothesis of the Extreme Value Theorem, and illustrate on your graph that the conclusion of the Extreme Value Theorem follows.

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