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Let $\left \lfloor {x} \right \rfloor$ to be floor value of $x$. $$ \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\arcsin \left\lfloor {(a + 2)x} \right\rfloor }}{{x + a}} - \cos \left\lfloor {\left| {ax} \right|} \right\rfloor } \right). $$ I have problem with this one. Help with direction how to solve it. It seems for me as any $a\neq 0$ is okay. As much as I understand cos we shoulв overlook as there is no uncertainty.

Gary
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qweasdzxcqweasd
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  • Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for. Here's a quick guide: https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question – 5xum May 04 '21 at 08:16
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    Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. Even if the question is closed, you can still edit it, and we will vote to reopen it. – 5xum May 04 '21 at 08:16
  • Can please someone give a hand? As much as I understand cos we shoulв overlook as there is no uncertaintyю – qweasdzxcqweasd May 04 '21 at 09:52
  • Someone will give a hand, once you edit your question according to my suggestions. – 5xum May 04 '21 at 10:33
  • In order for the limit to exist the left side limit must be equal to the right side limit meaning you approach 0 from -3,-2,-1 or from 3,2,1 – CuriousIndeed May 04 '21 at 11:45
  • How the left side limit look like? when a =0, right side - arcsin(0*x) ->0,cos(0) ->1. – qweasdzxcqweasd May 04 '21 at 13:48

1 Answers1

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The key observation is that $ \lfloor (a+2)x\rfloor $ and $\lfloor |ax| \rfloor$ both end up being constant for all sufficiently small $x$ with fixed sign, but the first ends at $-1$ or $0$ depending on the sign of $(a+2)x$. With that in mind, break down into cases:

  1. $a+2>0$ or $a+2 <0$. Then in the first instance, $$\lfloor(a+2)x\rfloor\to \left\{ \begin{array}{2}0, & \text{when } x \searrow 0 \\ -1, &\text{when } x \nearrow 0 \end{array}\right .$$ and in the second instance the limits are reversed, but continue to differ. So no limit can exist
  2. $a+2=0.$ Then $\lfloor(a+2)x\rfloor=0$ for all $x$ and $\lfloor|ax|\rfloor = 0$ for sufficiently small $x$. The limit then exists and is $\cos(0) =1 $.

Note that $a=0$ is covered in case 1, but it has an additional problem that when $\lfloor (a+2)x \rfloor \to -1$ (one of $x$ and $a+2$ is negative), the left term becomes infinite.

If you restrict $x$ to positive values only then a different conclusion is reached in case 1. If $(a+2)>0$ the left term eventually becomes zero and you have the same limit as case 2. If $(a+2)<0$ then the numerator of left term has limit $\arcsin(-1) = -\pi/2$. As long as $a\neq 0$ the left term then has limit $-\pi/(2a)$. The right term always has limit $1$.

WA Don
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