My Math Forum limit convergent

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 September 20th, 2018, 05:48 PM #11 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 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 05:52 PM.
September 21st, 2018, 03:14 AM   #12
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$\displaystyle \lim_{y\rightarrow 0} 1/y^{a}$ 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 |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

September 21st, 2018, 03:26 AM   #13
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Quote:
 Originally Posted by zylo $\displaystyle \lim_{y\rightarrow 0} 1/y^{a}$ 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 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.

September 21st, 2018, 04:08 AM   #14
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Quote:
 Originally Posted by v8archie 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.

 September 21st, 2018, 05:19 AM #15 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 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 05:29 AM.
September 21st, 2018, 05:27 AM   #16
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Quote:
 Originally Posted by v8archie 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.

 September 21st, 2018, 05:30 AM #17 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 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
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Quote:
 Originally Posted by SDK 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.

September 21st, 2018, 08:10 AM   #19
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Quote:
 Originally Posted by zylo 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.

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