coooool222
Junior Member
- Joined
- Jun 1, 2020
- Messages
- 93
I don't think this is a squeeze theorem problem!I am in dire confusion of this question. We have not learned derivatives yet, I am learning the squeeze thereom.
Here is the problem
View attachment 34147
Work:
-1 =< sin x/9 =<1
-1/x =< (sin x/9)/x =< 1/x
= undefined?????????
When x==>0 sin x / x = 1I don't think this is a squeeze theorem problem!
Have you learned the special limit [imath]\displaystyle\lim_{x\to0}\frac{\sin(x)}{x}[/imath]? That is usually taught as a fact to learn and apply.
You would have to change it toWhen x==>0 sin x / x = 1
Multiply it by 1/9 its 1/9?
Wouldn't that be sin x/9x?You would have to change it to
[imath]\dfrac{sin \left ( \dfrac{x}{9} \right ) }{ \left ( \dfrac{x}{9} \right ) } = 9 \cdot \dfrac{ sin \left ( \dfrac{x}{9} \right ) }{x}[/imath]
-Dan
It would be. Personally. I find nicely typeset formulae more helpful -- wonder if you find it more helpful too:Wouldn't that be sin x/9x?
It would be. Personally. I find nicely typeset formulae more helpful -- wonder if you find it more helpful too:
[math]\frac{\sin x}{9\cdot x}[/math]
Thank you for helping but can I also solve this question with squeeze thereom?[math]\frac{1}{9}*sin x/x \\ = 1/9[/math]
I don't know. Is this a requirement of your homework, or just a matter of personal curiosity?Thank you for helping but can I also solve this question with squeeze thereom?
No, it isn't really that. You could change variables, letting u=x/9, and it would be the limit of sin(u)/(9u). You'll need to learn how to properly show your work (and how to write your work carefully so that you can avoid mistakes).Wouldn't that be sin x/9x?
Can you show us your graph, and why it looks undefined to you? I suspect you may have a wrong idea about what makes a limit undefined (or else a wrong graph). Rather than defer to "sources", you should try to understand how things work (and why you first impressions are wrong).I tried graphing the equation and its telling me the limit is undefined. But then other sources are saying its 1/9.
Oh wait I forgot. If there is a discontinuity in a graph the limit is still there but the function itself isn't defined.No, it isn't really that. You could change variables, letting u=x/9, and it would be the limit of sin(u)/(9u). You'll need to learn how to properly show your work (and how to write your work carefully so that you can avoid mistakes).
Can you show us your graph, and why it looks undefined to you? I suspect you may have a wrong idea about what makes a limit undefined (or else a wrong graph). Rather than defer to "sources", you should try to understand how things work (and why you first impressions are wrong).
Yes, the function is undefined, but the limit is not. That's part of the reason limits were invented in the first place!Oh wait I forgot. If there is a discontinuity in a graph the limit is still there but the function itself isn't defined.
View attachment 34150
It's approaching 0.111 = 1/9.
If you multiply the limit by anything other than 1, this new limit will not be equal to the original limit. Why not multiply by (1/9)/(1/9)???When x==>0 sin x / x = 1
Multiply it by 1/9 its 1/9?
This equation is not valid at all. It seems that you are thinking that sinx/x = 1. This is not true at all![math]\frac{1}{9}*sin x/x \\ = 1/9[/math]
I meant the limit of sin x / x as x==> 0This equation is not valid at all. It seems that you are thinking that sinx/x = 1. This is not true at all!
... sin x / x ...
Steven G said:... sinx/x ...
coooool222 said:... sin x/9x...