Based on the example, state whether each statement is true or false.|x−2|<1 implies (C) |f(x)−1|<ϵ implies (D) (D) implies (C) (C) implies (D) (D) implies |f(x)−1|<ϵ complete.Example 2. Prove that lim f(r) = 1 by use of Definition 3, where f is given by2x +2if a 2f(エ) %=2if r = 2This function is almost identical to Example 2 on page 4 for pre-images, but with a noticeable difference: it has ajump at ro = 2, which is the main reason that I, (ro) \ {ro} occurs in the definition of limits,Proof. Let e>0 and let 1

Given the proof of limx→2fx=1, where fx=x2-2x+22if x≠22if x=2 Here equation (c) is, fx-1≤32x-2 Also,…

Answered: Example 2. Prove that lim f(r) =1 by…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Based on the example, state whether each statement is true or false.

\|x−2\|<1 implies (C)
\|f(x)−1\|<ϵ implies (D)
(D) implies (C)
(C) implies (D)
(D) implies \|f(x)−1\|<ϵ

complete.
Example 2. Prove that lim f(r) = 1 by use of Definition 3, where f is given by
2x +2
if a 2
f(エ) %=
2
if r = 2
This function is almost identical to Example 2 on page 4 for pre-images, but with a noticeable difference: it has a
jump at ro = 2, which is the main reason that I, (ro) \ {ro} occurs in the definition of limits,
Proof. Let e>0 and let 1 <I<3 with r 2. Simple algebra shows that
3
If(x) – 1| = – 2| <
(C)
2.
The right hand side of (C) being less than E on the domain 0< r 2 <l is equivalent to
2e
0<r 2 <8= min 1,
3
(D)
Hence f (1-6, 1+e) contains a perforated &-neighborhood of 2 defined by (D). The proof is complete.
21
A A V

Expert Solution

Step 1

Given the proof of $\lim_{x \to 2} f (x) = 1$ , where $f (x) = \{\begin{cases} \frac{x^{2} - 2 x + 2}{2} & if x \neq 2 \\ 2 & if x = 2 \end{cases}$

Here equation (c) is, $|f (x) - 1| \leq \frac{3}{2} |x - 2|$

Also, equation (D) is, $0 < |x - 2| < δ = \min \{1, \frac{2 ε}{3}\}$

Now consider the first statement, $|x - 2| < 1$ does not imply equation (C).

$\begin{array}{rcl} |f (x) - 1| & = & |\frac{x^{2} - 2 x + 2}{2} - 1| \\ = & |\frac{x^{2} - 2 x + 2 - 2}{2}| \\ = & \frac{|x|}{2} |x - 2| \\ \leq & \frac{3}{2} |x - 2| [∵ x < 3] \end{array}$

So, $|x - 2| < 1$ does not imply equation (C).

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