Squeeze theorem problem

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coooool222

Junior Member
Joined
Jun 1, 2020
Messages
93
I am in dire confusion of this question. We have not learned derivatives yet, I am learning the squeeze thereom.
Here is the problem

1664253504533.png
Work:
-1 =< sin x/9 =<1
-1/x =< (sin x/9)/x =< 1/x
= undefined?????????
 
coooool222 said:
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?????????
Click to expand...
I 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.
 
Dr.Peterson said:
I 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.
Click to expand...
When x==>0 sin x / x = 1
Multiply it by 1/9 its 1/9?
 
coooool222 said:
When x==>0 sin x / x = 1
Multiply it by 1/9 its 1/9?
Click to expand...
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
 
topsquark said:
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
Click to expand...
Wouldn't that be sin x/9x?

I tried graphing the equation and its telling me the limit is undefined. But then other sources are saying its 1/9.
 
coooool222 said:
Wouldn't that be sin x/9x?
Click to expand...
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).
coooool222 said:
I tried graphing the equation and its telling me the limit is undefined. But then other sources are saying its 1/9.
Click to expand...
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).
 
Dr.Peterson said:
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).
Click to expand...
Oh wait I forgot. If there is a discontinuity in a graph the limit is still there but the function itself isn't defined.
1664299091770.png
It's approaching 0.111 = 1/9.
 
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