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September 19th, 2018, 08:12 PM   #1
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limit convergent

]How can I approach this question, please?
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 September 19th, 2018, 08:53 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,600 Thanks: 2588 Math Focus: Mainly analysis and algebra 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 Thanks from topsquark and Leonardox Last edited by v8archie; September 19th, 2018 at 08:58 PM.
 September 19th, 2018, 11:23 PM #3 Senior Member   Joined: Apr 2017 From: New York Posts: 119 Thanks: 6 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?
 September 19th, 2018, 11:38 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,600 Thanks: 2588 Math Focus: Mainly analysis and algebra 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 $a  September 20th, 2018, 06:31 AM #5 Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 |sinx|$\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$0 Thanks from Leonardox September 20th, 2018, 09:11 AM #6 Senior Member Joined: Sep 2016 From: USA Posts: 562 Thanks: 325 Math Focus: Dynamical systems, analytic function theory, numerics Quote:  Originally Posted by zylo |sinx|$\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$0 This is not correct. A simple counterexample is$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$y \to 0$. September 20th, 2018, 12:28 PM #7 Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 Quote:  Originally Posted by SDK 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. September 20th, 2018, 02:30 PM #8 Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 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$0 Except a=0 and b=0, in which case OP limit is sin(1). September 20th, 2018, 03:59 PM #9 Senior Member Joined: Sep 2016 From: USA Posts: 562 Thanks: 325 Math Focus: Dynamical systems, analytic function theory, numerics Quote:  Originally Posted by zylo For a=-1, 1/y$\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$. I am sure you are just trolling as usual but do that in your own nonsense threads. You are just confusing someone else who is asking for help. September 20th, 2018, 05:31 PM #10 Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 Quote:  Originally Posted by zylo For a=-1, 1/y$\displaystyle ^{a}\$ = y whose limit is zero.

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