Chapter 1: Q. 18 (page 98)

Show that the limit as $x \to 2 of f (x) = \sqrt{x - 1.1}$ is not equal to 1, by finding an $ε > 0$ for which there is no corresponding $δ > 0$ satisfying the formal definition of limit.

Short Answer

Expert verified

The $δ < 0$ , the expression is false.

Step by step solution

Step 1. Given information.

The given function is $\lim_{x \to 2} \sqrt{x - 1.1} \neq 1$ .

Step 2. Calculation.

From the given expression, we have, $c = 2, L = 1$ .

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

For all epsilon positive, there exists a delta positive such that if,

$x \in (2 - δ, 2) \cup (2, 2 + δ) then \sqrt{x - 1.1} \in (1 - ε, 1 + ε)$ .

So the largest value of $δ$ is:

$δ = (1^{2} - 1.1) - 2 δ = 1 - 1.1 - 2 δ = 1 - 3.1 δ = - 2.1$

Since $δ < 0$ , the expression is false.

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Show that the limit as $x \to 2 of f (x) = \sqrt{x - 1.1}$ is not equal to 1, by finding an $ε > 0$ for which there is no corresponding $δ > 0$ satisfying the formal definition of limit.

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Calculation.

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Step by step solution

Step 1. Given information.

Step 2. Calculation.

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Show that the limit as $x \to 2 of f (x) = \sqrt{x - 1.1}$ is not equal to 1, by finding an $ε > 0$ for which there is no corresponding $δ > 0$ satisfying the formal definition of limit.