My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree6Thanks
Reply
 
LinkBack Thread Tools Display Modes
September 20th, 2018, 06:48 PM   #11
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,515
Thanks: 2515

Math Focus: Mainly analysis and algebra
For negative $b=-c$ (where $c>0$) we have $\frac1{y^b}=\frac1{y^{-c}}=y^c \to 0$. We can write $y^a= \frac{y^{a+c}}{y^c}$ and the limit is now
$$\lim_{y \to 0^+} y^{a+c} \cdot \frac{\sin y^c}{y^c} = \lim_{y \to 0^+} y^{a+c}$$
This limit exists for all non-negative values of $(a+c)$ so we have $$a+c \ge 0 \implies a \ge -c \implies a \ge b$$

A non-trivial example of this is $a=-2$, $b=-3$.

Last edited by v8archie; September 20th, 2018 at 06:52 PM.
v8archie is offline  
 
September 21st, 2018, 04:14 AM   #12
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,603
Thanks: 115

I find your threads incomprehensible, and SDK's nasty in addition to being nonsense.

$\displaystyle \lim_{y\rightarrow 0} 1/y^{a}$ is either 0 (a<0), or $\displaystyle \infty$ (a>0).

if a=0, $\displaystyle \lim_{y\rightarrow 0} 1/y^{0}$ is 1 and sin term oscillates for all b (no convergence) unless b=0, in which case limit is 1xsin1 = sin 1.


Reference, but note previous correction for a=0:

Quote:
Originally Posted by zylo View Post
|sinx|$\displaystyle \le$1, all x (all b)

let y = 1/n

lim 1/y$\displaystyle ^{a}$ = lim n$\displaystyle ^{a}$ = $\displaystyle \infty$ for a>0, = 0 for all a $\displaystyle \le$ 0
zylo is offline  
September 21st, 2018, 04:26 AM   #13
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,515
Thanks: 2515

Math Focus: Mainly analysis and algebra
Quote:
Originally Posted by zylo View Post
$\displaystyle \lim_{y\rightarrow 0} 1/y^{a}$ is either 0 (a<0), or $\displaystyle \infty$ (a>0).
True, but it's not the whole story. There's a second term there.

Quote:
Originally Posted by zylo View Post
if a=0, $\displaystyle \lim_{y\rightarrow 0} 1/y^{0}$ is 1 and sin term oscillates for all b (no convergence)
This is false, for example $\displaystyle \lim_{x \to 0} \sin x$.

I will grant that my solution is incomplete (I never intended to give a complete solution), but I'm surprised that you find it incomprehensible. I think very carefully about the clarity of my style. Please tell me what part of my post you can't follow, I'll endeavour to clarify.
v8archie is offline  
September 21st, 2018, 05:08 AM   #14
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,603
Thanks: 115

Quote:
Originally Posted by v8archie View Post
True, but it's not the whole story. There's a second term there.


*This is false, for example $\displaystyle \lim_{x \to 0} \sin x$.

I will grant that my solution is incomplete (I never intended to give a complete solution), but I'm surprised that you find it incomprehensible. I think very carefully about the clarity of my style. Please tell me what part of my post you can't follow, I'll endeavour to clarify.
There is a second term (sin) which is irrelevant (unless a=0 and b=0) since its absolute value is $\displaystyle \le$1 and $\displaystyle \lim_{y\rightarrow 0} 1/y^{a}$ is 0 or $\displaystyle \infty$, unless a=0.

*I wrote " if a=0, $\displaystyle \lim_{y\rightarrow 0} 1/y^{0}$ is 1 and sin term oscillates for all b (no convergence) unless b=0, in which case limit is 1xsin1 = sin 1."
Correction: sin(1/y$\displaystyle ^{b}$) only oscillates if b > 0, and approaches 0 if b < 0. But it has no influence on convergence unless b=0 and a=0.

I fail to see the point of c. The sin term simply doesn't matter unless a=0 and b=0, in which case convergence is to sin1.
zylo is offline  
September 21st, 2018, 06:19 AM   #15
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,515
Thanks: 2515

Math Focus: Mainly analysis and algebra
But it's not irrelevant. Or are you denying that $\displaystyle \lim_{x \to 0} \frac{\sin x}x = 1$? Just look at the graph of $x^{-2}\sin \frac1{x^{-3}}$!

The $c$ is there because variables with negative values can confuse.

You seem to be stuck in the position you commonly find yourself in. You've made an assumption which it turns out is unjustified, but rather than admit any error on your part you will defend your erroneous position disregarding any and all evidence to the contrary - possibly over the course of several threads.
Thanks from greg1313

Last edited by v8archie; September 21st, 2018 at 06:29 AM.
v8archie is offline  
September 21st, 2018, 06:27 AM   #16
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,603
Thanks: 115

Quote:
Originally Posted by v8archie View Post
But it's not irrelevant. Or are you denying that $\displaystyle \lim_{x \to 0} \frac{\sin x}x = 1$?
No. But neither do I deny that 2+2=4. It's irrelevant and I explained why.

OP can make his own choice, this is getting tedious.
zylo is offline  
September 21st, 2018, 06:30 AM   #17
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,515
Thanks: 2515

Math Focus: Mainly analysis and algebra
See my edit.

You're right, your intransigence is very tedious. It's why your relationship with various people here is so strained.
v8archie is offline  
September 21st, 2018, 08:04 AM   #18
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,885
Thanks: 1088

Math Focus: Elementary mathematics and beyond
Quote:
Originally Posted by SDK View Post
Obviously, you are even willing to double down when wrong about simple arithmetic . $\frac{1}{y} \to \infty$ as $y \to 0$. I am sure you are just trolling as usual but do that in your own nonsense threads. You are just confusing someone else who is asking for help.
I'm willing to give zylo the benefit of the doubt.

zylo, please give the OP a chance to respond (24 hrs) before you implement your own agenda.
greg1313 is offline  
September 21st, 2018, 09:10 AM   #19
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,515
Thanks: 2515

Math Focus: Mainly analysis and algebra
Quote:
Originally Posted by zylo View Post
OP can make his own choice
The OP came looking for help in getting the right answer. It is polite to admit your error and withdraw your incorrect solution when it is pointed out.

I encourage the OP to at least look at the graph I posted before choosing which method is correct.
v8archie is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
convergent, limit



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
two convergent sequences limit of their product annakar Real Analysis 0 December 9th, 2012 02:04 AM
convergent rose3 Real Analysis 1 December 6th, 2009 12:19 AM
convergent rose3 Real Analysis 1 December 2nd, 2009 02:25 PM
convergent rose3 Real Analysis 1 November 19th, 2009 10:12 AM
Find convergent subsequence of a non convergent sequence boxerdog246 Real Analysis 1 October 5th, 2008 01:57 AM





Copyright © 2018 My Math Forum. All rights reserved.