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Chapter 1: Q. 13 (page 135)

Suppose f and g are functions such that $\lim_{x 鈫� 3} f (x) = 5, \lim_{x 鈫� 4} f (x) = 2$ and $\lim_{x 鈫� 3} g (x) = 4$
Given this information, calcuate the limits that follow, if possible. If it is not possible with the given information, explain why.
$\lim_{x 鈫� 3} (2 f (x) - 3 g (x))$

Short Answer

Expert verified

The value of $\lim_{x 鈫� 3} (2 f (x) - 3 g (x)) = - 2$

Step by step solution

01

Step 1. Given information

The given limits $\lim_{x 鈫� 3} (2 f (x) - 3 g (x))$ and the given values are,

$\lim_{x 鈫� 3} f (x) = 5, \lim_{x 鈫� 4} f (x) = 2$ and $\lim_{x 鈫� 3} g (x) = 4$

02

Step 2. Calculating value of the given limits

The value can be calculated as below,

$\lim_{x 鈫� 3} (2 f (x) - 3 g (x)) = \lim_{x 鈫� 3} (2 f (x)) - \lim_{x 鈫� 3} (3 g (x)) [s u m r u l e] = 2 \lim_{x 鈫� 3} f (x) - 3 \lim_{x 鈫� 3} g (x) [c o n s \tan t m u l t i p l e r u l e] = 2 (5) - 3 (4) = 10 - 12 = - 2$

Hence,the result is $- 2$

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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| < $蔚$ .

$\lim_{x 鈫� 3} (x^{2} - 6 x + 5) = - 4$

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 $x 鈭� (c - 未, c) 鈭� (c, c + 未) then f (x) 鈭� (L - 蔚, L + 蔚)$

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

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

$\lim_{x 鈫� 2^{+}} 3 \sqrt{2 x - 4} = 0$

Write each of the inequalities in interval notation:

$|(3 x + 1) - 2| < 0.1$

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.

$f (x) = x^{鈭� 1}$

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