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September 19th, 2018, 07:12 PM   #1
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limit convergent

]How can I approach this question, please?
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September 19th, 2018, 07:53 PM   #2
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Notice that as $y \to 0^+$ the expression $\frac1{y^b}$ grows without bound for all $b \gt 0$. In this case the sine function oscillates more and more rapidly, so you need the other term to shrink the oscillations to zero.

When $b = 0$, the sine function is constant. You can consider what that means for the other term.

When $b < 0$, $\frac1{y^b} \to 0$ and so you will need to consider that $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$
By setting $x = \frac1{y^b}$, you will be able to reduce the given limit to one that only involves powers of $y$ and thus solve the problem
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Last edited by v8archie; September 19th, 2018 at 07:58 PM.
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September 19th, 2018, 10:23 PM   #3
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So to be able to graph on a and b axes
what values should I give for a and b
or how can I draw it?
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September 19th, 2018, 10:38 PM   #4
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For each of those three cases you should be able to determine the boundary between convergence and divergence in the form of a curve. For example if it is required that $a<b$, the boundary is $a=b$. You can draw this and shade the appropriate side.
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September 20th, 2018, 05:31 AM   #5
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|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
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September 20th, 2018, 08:11 AM   #6
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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
This is not correct. A simple counterexample is $a = -1$. v8archie has the correct idea. Another way to approach it which is similar to his separation into 3 cases is to expand $\sin x$ for $x \approx 0$ and apply the mean value theorem. This gives an expression of the form
\[ \sin(x) = x\cos(\theta x + 1 - \theta) \ \text{for some } \theta \in (0,1) \]
Evaluate at $y^{-b}$, multiply by $y^{a}$ and let $y \to 0$.
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September 20th, 2018, 11:28 AM   #7
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Quote:
Originally Posted by SDK View Post
This is not correct. A simple counterexample is $a = -1$.
For a=-1, 1/y$\displaystyle ^{a}$ = y whose limit is zero (given).

Your conclusion that I was not correct based on an incorrect counterexample is wrong.

Presumably (hopefully) OP will eventually learn correct answer in class. Considering his judgement in awarding thanks, I would appreciate it if he would remove mine.
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September 20th, 2018, 01:30 PM   #8
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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
Except a=0 and b=0, in which case OP limit is sin(1).
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September 20th, 2018, 02:59 PM   #9
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Quote:
Originally Posted by zylo View Post
For a=-1, 1/y$\displaystyle ^{a}$ = y whose limit is zero (given).

Your conclusion that I was not correct based on an incorrect counterexample is wrong.

Presumably (hopefully) OP will eventually learn correct answer in class. Considering his judgement in awarding thanks, I would appreciate it if he would remove mine.
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.
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September 20th, 2018, 04:31 PM   #10
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Quote:
Originally Posted by zylo View Post
For a=-1, 1/y$\displaystyle ^{a}$ = y whose limit is zero.
Can't follow that?
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