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

[huan.conchito]

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

thank you very much

 

[dextercioby]

Okay.

\lim_{x\rightarrow +\infty} \frac{\sin \frac{1}{x}}{\frac{1}{\sqrt{x}}}

Use the limit

\lim_{y\searrow 0} \frac{\sin y}{y}=1

and the substitution

\frac{1}{x}=y

Daniel.

 

[snoble]

hmm... how about a slightly easier way. We have that
x \ge 0 \Rightarrow sin(x) \le x
so
lim_{x\rightarrow \infty} \sqrt{x}\times sin(1/x) \le lim_{x\rightarrow \infty} \sqrt{x}/x=?
Now can you show whether or not the limit is positive? What about negative?

 

[huan.conchito]

so its = 0?

 

[dextercioby]

That way u can show the original limit is smaller or less than 0.U need the "squeeze theorem"...,U can use

x > 0 \Rightarrow -x <\sin x< x

to start with,and the afore mentioned theorem

Daniel.

 

[snoble]

I don't think I understand your complaint Daniel. All I'm saying is sin(1/x) \le 1/x when x is positive. How does that involve the square root?

 

[dextercioby]

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.