Limit of SQRT(X)SIN(1/X) as X Goes to Infinity

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Discussion Overview

The discussion centers on evaluating the limit of the expression SQRT(X)SIN(1/X) as X approaches infinity. Participants explore different approaches to determine the limit, including the use of known limits and inequalities.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes using the limit of sin(y)/y as y approaches 0 to evaluate the limit of SQRT(X)SIN(1/X) as X approaches infinity.
  • Another participant suggests an alternative approach using inequalities, noting that sin(x) is less than or equal to x, and questions whether the limit is positive or negative.
  • A participant expresses uncertainty about the implications of the inequalities and the application of the squeeze theorem in this context.
  • There is a clarification regarding the relationship between sin(1/x) and 1/x, with acknowledgment of the need for the squeeze theorem to establish the limit.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the limit's value, and multiple approaches and interpretations are presented, indicating ongoing debate and uncertainty.

Contextual Notes

Participants rely on various mathematical theorems and inequalities, but there are unresolved aspects regarding the application of the squeeze theorem and the implications of the inequalities discussed.

huan.conchito
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Lim x-> infinity of
SQRT(X)SIN(1/X)

thank you very much
 
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Okay.

[tex]\lim_{x\rightarrow +\infty} \frac{\sin \frac{1}{x}}{\frac{1}{\sqrt{x}}}[/tex]

Use the limit

[tex]\lim_{y\searrow 0} \frac{\sin y}{y}=1[/tex]

and the substitution

[tex]\frac{1}{x}=y[/tex]

Daniel.
 
hmm... how about a slightly easier way. We have that
[tex]x \ge 0 \Rightarrow sin(x) \le x[/tex]
so
[tex]lim_{x\rightarrow \infty} \sqrt{x}\times sin(1/x) \le lim_{x\rightarrow \infty} \sqrt{x}/x=?[/tex]
Now can you show whether or not the limit is positive? What about negative?
 
so its = 0?
 
That way u can show the original limit is smaller or less than 0.U need the "squeeze theorem"...,U can use

[tex]x > 0 \Rightarrow -x <\sin x< x[/tex]

to start with,and the afore mentioned theorem

Daniel.
 
Last edited:
I don't think I understand your complaint Daniel. All I'm saying is [tex]sin(1/x) \le 1/x[/tex] when x is positive. How does that involve the square root?
 
Yeah,you're right.I was thinking of my post and the substitution involved there.Yes,those 2 inequalities can prove it,using "squeeze theorem".

Daniel.
 

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