Can you help please!Consider the function f(x) = x3 for x ∈ ℝ and –2 < x < 4. Prove that limx→1 f(x) = 1. Use the method shown in Example 3.7.5. (I attached Example 3.7.5 files) 2x2-5x-3Vx (0 < x - 3|< 8 →x-3-7| 0380Vx (0 < |x − 3] < 8 →| (2x2-5x-3)/(x − 3) — 7| < €).--- Example 3.7.5. Show that2x2-5x-3limx→3 x-3= 7.Scratch workAccording to the definition of limits, our goal means that for every positivenumber € there is a positive number 8 such that if x is any number such that0<|x3|< 8, then | (2x² -5x-3)/(x-3) - 7| < €. Translating this intological symbols, we haveVe>0x0<2x2-5x-3-31 < 8→x-3기).< EWe therefore start by letting & be an arbitrary positive number and then try tofind a positive number 8 for which we can proveVx0<|x3| < 8 →|2x2-5x-3x-3-7|

We have to prove that limx→1f(x)=1. To do that, we need to show that for any given ϵ>0, there…

Answered: Consider the function f(x) = x3 for x ∈…

Math

Advanced Math

Consider the function f(x) = x3 for x ∈ ℝ and –2 < x < 4. Prove that limx→1 f(x) = 1

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

Can you help please!

Consider the function f(x) = x³ for x ∈ ℝ and –2 < x < 4. Prove that lim_x→1 f(x) = 1. Use the method shown in Example 3.7.5. (I attached Example 3.7.5 files)

2x2-5x-3
Vx (0 < x - 3|< 8 →
x-3
-7| <e).
The scratch work involved in finding 8 will not appear in the proof, of course.
In the final proof we'll just write "Let 8 = (some positive number)" and then
proceed to prove
Vx
(0 < x - 3|< 8 →
|2x2-5x-3
x-3
3 -7| < €).
Before working out the value of 8, let's figure out what the rest of the proof
will look like. Based on the form of the goal at this point, we should proceed
by letting x be arbitrary, assuming 0 < |x − 3] < 8, and then proving | (2x² -
5x-3)/(x-3)-7| < €. Thus, the entire proof will have the following form:
Let € be an arbitrary positive number.
Let 8 = (some positive number).
Let x be arbitrary.
Suppose 0 < x— 3| < 8.
-
[Proof of | (2x2-5x-3)/(x-3) - 7| < € goes here.]
-
Therefore 0 < x - 3|<8 → |(2x² - 5x-3)/(x-3)-7| < €.
Since x was arbitrary, we can conclude that Vx(0 < |x − 3] < 8 →
|(2x2-5x-3)/(x − 3) — 7| < €).
-
-
Therefore 380Vx(0 < |x −3] < 8 → |(2x²-5x-3)/(x-3)−7| < €).
Since & was arbitrary, it follows that Ve > 0380Vx (0 < |x − 3] < 8 →
| (2x2-5x-3)/(x − 3) — 7| < €).
-
-
-

Example 3.7.5. Show that
2x2-5x-3
lim
x→3 x-3
= 7.
Scratch work
According to the definition of limits, our goal means that for every positive
number € there is a positive number 8 such that if x is any number such that
0<|x3|< 8, then | (2x² -5x-3)/(x-3) - 7| < €. Translating this into
logical symbols, we have
Ve>0x0<
2x2-5x-3
-31 < 8→
x-3
기).
< E
We therefore start by letting & be an arbitrary positive number and then try to
find a positive number 8 for which we can prove
Vx0<|x3| < 8 →
|2x2-5x-3
x-3
-7|<e).
The scratch work involved in finding 8 will not appear in the proof, of course.
In the final proof we'll just write "Let 8 = (some positive number)" and then
proceed to prove
Vx
5/2r2-5x-3
-7|<e).
<€).
(0 < x - 3|< 8 →
Before working out the value of 8, let's figure out what the rest of the proof
will look like. Based on the form of the goal at this point, we should proceed
by letting x be arbitrary, assuming 0 < |x − 3] < 8, and then proving | (2x² -
5x-3)/(x-3)-7] < €. Thus, the entire proof will have the following form:
Let € be an arbitrary positive number.
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