September 20th, 2018, 05:48 PM  #11 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 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 nonnegative values of $(a+c)$ so we have $$a+c \ge 0 \implies a \ge c \implies a \ge b$$ A nontrivial example of this is $a=2$, $b=3$. Last edited by v8archie; September 20th, 2018 at 05:52 PM. 
September 21st, 2018, 03:14 AM  #12 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,529 Thanks: 107 
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: 
September 21st, 2018, 03:26 AM  #13  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra  Quote:
Quote:
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.  
September 21st, 2018, 04:08 AM  #14  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,529 Thanks: 107  Quote:
*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.  
September 21st, 2018, 05:19 AM  #15 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 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. Last edited by v8archie; September 21st, 2018 at 05:29 AM. 
September 21st, 2018, 05:27 AM  #16 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,529 Thanks: 107  
September 21st, 2018, 05:30 AM  #17 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 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. 
September 21st, 2018, 07:04 AM  #18  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,878 Thanks: 1087 Math Focus: Elementary mathematics and beyond  Quote:
zylo, please give the OP a chance to respond (24 hrs) before you implement your own agenda.  
September 21st, 2018, 08:10 AM  #19 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra  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. 

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