Chapter 1: Q. 54 (page 98)

For each limit $\underset{x \to c}{l i m} f (x) = L$ in Exercises 43–54, use graphs and algebra to approximate the largest value of $δ$ such that if $x \in (c - δ, c) \cup (c, c + δ) then f (x) \in (L - ε, L + ε)$
$\underset{x \to - 1}{l i m} \frac{x^{2} - 2 x - 3}{x + 1} = - 4, ε = 0.1$

Short Answer

Expert verified

The required value of $δ = 0.1$

Step by step solution

Step 1. Given Information

The given expression is $\underset{x \to - 1}{l i m} \frac{x^{2} - 2 x - 3}{x + 1} = - 4, ε = 0.1$

Step 2. Explanation

From the given expression, we have, $c = - 1, L = - 4$

The limit expression can be written as a formal statement as below,

For all epsilon positive, there exists a delta positive such that if $x \in (- 1 - δ, - 1) \cup (- 1, - 1 + δ) Then \frac{x^{2} - 2 x - 3}{x + 1} \in (- 4 - ε, - 4 + ε)$

Now, the largest value of delta is given by,

$\frac{x^{2} - 2 x - 3}{x + 1} = L + ε \frac{x^{2} - 2 x - 3}{x + 1} = - 4 + 1 x - 3 = - 3 x = 0$

Thus,

$δ = 0 + 0.1 δ = 0.1$

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For each limit $\underset{x \to c}{l i m} f (x) = L$ in Exercises 43–54, use graphs and algebra to approximate the largest value of $δ$ such that if $x \in (c - δ, c) \cup (c, c + δ) then f (x) \in (L - ε, L + ε)$
$\underset{x \to - 1}{l i m} \frac{x^{2} - 2 x - 3}{x + 1} = - 4, ε = 0.1$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Explanation

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For each limit limx→cf(x)=Lin Exercises 43–54, use graphs and algebra to approximate the largest value of δsuch that ifx∈(c-δ,c)∪(c,c+δ)thenf(x)∈(L-ε,L+ε)limx→-1x2-2x-3x+1=-4,ε=0.1

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Explanation

One App. One Place for Learning.

Most popular questions from this chapter

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For each limit $\underset{x \to c}{l i m} f (x) = L$ in Exercises 43–54, use graphs and algebra to approximate the largest value of $δ$ such that if $x \in (c - δ, c) \cup (c, c + δ) then f (x) \in (L - ε, L + ε)$
$\underset{x \to - 1}{l i m} \frac{x^{2} - 2 x - 3}{x + 1} = - 4, ε = 0.1$