This is true if and only if |x-2|< 8 = 5=53₁ 1b) We have In our case, domf = R\ {0}. if and only if, for any RE R, there exists > 0, such that, for any x € domf with 1 0< x-1| <8, we have that 1 lim x1 (x-1)2 This is equivalent to 1 lim x+1(x-1)2 we fix R> 0 and we consider the condition that is > R. (x - 1)² To verify that = +∞ f(x) = = +∞ 1 (x - 1)² (x - 1)² < 1/1/2 > R. 1 |x-1|< 8 = √√R

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
icon
Related questions
Question
Exercise 1b) I have attached the solution below The part I don’t understand is why did they write dom f= R /{0} shouldn’t it be R /{1} correct me if I’m wrong
Limits of functions with standard
techniques: solved exercises
Exercise 1. Using the definition of limit, verify that
(a) lim (3x - 5) = 1
x-2
(c)
3x²
lim
x+∞ x + 1
= +∞
(b)
(d)
an
1
lim
x-1(x - 1)²
3x + 1
lim
x-xx-2
Exercise 3. Using the definition of limit, verify that the
2n + 3
3n - -7
Exercise 2. Let a be a real positive number. Prove that if n E N and n > [a], then
n>a (where [a] denotes the floor (or integer part) of a).
+∞
= 3
sequence
Transcribed Image Text:Limits of functions with standard techniques: solved exercises Exercise 1. Using the definition of limit, verify that (a) lim (3x - 5) = 1 x-2 (c) 3x² lim x+∞ x + 1 = +∞ (b) (d) an 1 lim x-1(x - 1)² 3x + 1 lim x-xx-2 Exercise 3. Using the definition of limit, verify that the 2n + 3 3n - -7 Exercise 2. Let a be a real positive number. Prove that if n E N and n > [a], then n>a (where [a] denotes the floor (or integer part) of a). +∞ = 3 sequence
€
This is true if and only if |x-2|< 6 =
16) We have
In our case, domf = R\ {0}.
if and only if, for any RE R, there exists > 0, such that, for any x E domf with
1
0< x-1| <8, we have that
> R.
(x - 1)²
To verify that
1
lim
x1 (x-1)2
1
lim
21 (x-1)2
we fix R> 0 and we consider the condition
This is equivalent to
that is
= +∞
f(x) =
= +∞
1
(x - 1)²
(x - 1)² < 1/1/2
> R.
1
|x-1|< 8 = √²
Transcribed Image Text:€ This is true if and only if |x-2|< 6 = 16) We have In our case, domf = R\ {0}. if and only if, for any RE R, there exists > 0, such that, for any x E domf with 1 0< x-1| <8, we have that > R. (x - 1)² To verify that 1 lim x1 (x-1)2 1 lim 21 (x-1)2 we fix R> 0 and we consider the condition This is equivalent to that is = +∞ f(x) = = +∞ 1 (x - 1)² (x - 1)² < 1/1/2 > R. 1 |x-1|< 8 = √²
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,