Does the lim x→0 of (x^4)(sin1/x) exist?

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  • #1
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Homework Statement


I need to solve this limit.
lim (x^4)(sin1/x)
x→0

Homework Equations





The Attempt at a Solution



Since as 1/x → 0+, the limit is +∞ but when 1/x → 0-, the limit is -∞. Since these two-sided limits don't match I thought the limit could not be calculated. apparently this is incorrect. any thoughts?
 

Answers and Replies

  • #2
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Yes it does exist. Basically you have to rearrange the function (I know I'm not using great math vocab here).

You take x^4*sin(1/x) and make it x^4 / (1/sin(1/x)). you then have 0/1...which the limit is 0.

Correct me if I am way off but that is how I would do it.
 
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  • #3
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The limit as x approaches 0 of sin(1/x) does not exist, so you can't use the usual theorems about products (or quotients) of limits.

But since [itex]|\sin x| \leq 1[/itex],
[tex]|x^4 \sin(1/x)| = |x^4| \cdot |\sin(1/x)| \leq |x^4| \cdot 1[/tex]
so the limit does exist and is zero.
 
  • #4
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The limit as x approaches 0 of sin(1/x) does not exist, so you can't use the usual theorems about products (or quotients) of limits.

But since [itex]|\sin x| \leq 1[/itex],
[tex]|x^4 \sin(1/x)| = |x^4| \cdot |\sin(1/x)| \leq |x^4| \cdot 1[/tex]
so the limit does exist and is zero.
Yep, thats right.
 
  • #5
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awesome. thanks guys!
 
  • #6
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Since as 1/x → 0+, the limit is +∞ but when 1/x → 0-, the limit is -∞. Since these two-sided limits don't match I thought the limit could not be calculated. apparently this is incorrect. any thoughts?
The problem isn't asking about 1/x. It's asking about sin(1/x), and in particular, the it is asking about the product of x4 with the sine of 1/x. You are correct in the sense that sin(1/x) is undefined as x→0. It is however bounded: [itex]-1\le \sin(1/x) \le 1[/itex] for all x. Thus [itex]-x^4\le x^4\sin(1/x)≤x^4[/itex]. What does the squeeze theorem say about this as x→0?
 
  • #7
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that since the limits on both of those sides equals 0 then the limit must also equal 0. that is extremely helpful. Thank you!
 

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