Simple Limit question

1. Apr 12, 2005

huan.conchito

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

thank you very much

2. Apr 12, 2005

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.

3. Apr 12, 2005

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?

4. Apr 12, 2005

huan.conchito

so its = 0?

5. Apr 12, 2005

dextercioby

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

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

Daniel.

Last edited: Apr 12, 2005
6. Apr 12, 2005

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?

7. Apr 12, 2005

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.