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Lim x -> inf. of sinx/x

  • Thread starter PhizKid
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  • #1
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Homework Statement


[tex]\lim_{x \to \infty } \frac{\sin x}{x}[/tex]


Homework Equations





The Attempt at a Solution


[tex]\lim_{x \to \infty } \frac{\sin x}{x}[/tex]
Substitute x = 1/u
[tex]\lim_{\frac{1}{u} \to 0 } \frac{\sin \frac{1}{u}}{\frac{1}{u}} = 1[/tex]

I know this is incorrect because intuitively I know that sinx will oscillate between 0 and 1, but the denominator will grow larger infinitely, therefore should theoretically approach 0 but that's just my thinking and no hard evidence.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Homework Statement


[tex]\lim_{x \to \infty } \frac{\sin x}{x}[/tex]


Homework Equations





The Attempt at a Solution


[tex]\lim_{x \to \infty } \frac{\sin x}{x}[/tex]
Substitute x = 1/u
[tex]\lim_{\frac{1}{u} \to 0 } \frac{\sin \frac{1}{u}}{\frac{1}{u}} = 1[/tex]

I know this is incorrect because intuitively I know that sinx will oscillate between 0 and 1, but the denominator will grow larger infinitely, therefore should theoretically approach 0 but that's just my thinking and no hard evidence.
Translate what you correctly know into an argument using the squeeze theorem.
 
  • #3
475
1
We haven't learned the squeeze theorem, so should I just take sinx/x as x-> inf as 0 for granted and just memorize it like I do with sinx/x as x-> 0? (We were given sinx/x as x-> 0 to be 1, but were told to just memorize it)
 
  • #4
Dick
Science Advisor
Homework Helper
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We haven't learned the squeeze theorem, so should I just take sinx/x as x-> inf as 0 for granted and just memorize it like I do with sinx/x as x-> 0? (We were given sinx/x as x-> 0 to be 1, but were told to just memorize it)
No, there's no need to memorize it. Remember -1<=sin(x)<=1, so -1/x<=sin(x)/x<=1/x. Even if you don't have the theorem that should make it pretty obvious that the limit is 0.
 
  • #5
Mentallic
Homework Helper
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[tex]\lim_{x \to \infty } \frac{\sin x}{x}[/tex]
Substitute x = 1/u
[tex]\lim_{\frac{1}{u} \to 0 } \frac{\sin \frac{1}{u}}{\frac{1}{u}} = 1[/tex]
That doesn't look right, if you're going to plug 1/u in for x then everything else needs to stay the same. As you have it there, you've changed the limit from [itex]x\to\infty[/itex] to [itex]\frac{1}{u} \to 0[/itex] when it should be [itex]\frac{1}{u} \to \infty[/itex]
But regardless, this won't help you.

We haven't learned the squeeze theorem, so should I just take sinx/x as x-> inf as 0 for granted and just memorize it like I do with sinx/x as x-> 0? (We were given sinx/x as x-> 0 to be 1, but were told to just memorize it)
Well then I guess you can just memorize it, but the squeeze theorem isn't very hard to understand really, especially for this example.

It's true that [tex]-1\leq \sin(x)\leq 1[/tex] right?
So if we divide all through by x, we get [tex]\frac{-1}{x}\leq \frac{\sin(x)}{x}\leq \frac{1}{x}[/tex]
And now we know that the limit as [itex]x\to\infty[/itex] for both [itex]-1/x[/itex] and [itex]1/x[/itex] is equal to zero, and since [itex]\sin(x)/x[/itex] is in between those, its limit must also be zero.
 
  • #6
475
1
Wow, thank you. That helps a ton, so now I know how to explicitly state this
 

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