Is Xsin 1 x Uniformly Continuous on 0 1

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Limit of xsin(1/x) as x→0

• Thread starter PirateFan308
• Start date Dec 17, 2011

Homework Statement

[itex]\displaystyle\lim_{x\rightarrow 0}~~ {xsin(1/x)}[/itex]

The Attempt at a Solution

I've attempted to solve this limit two different ways and get different answers.

Attempt #1:
If [itex](x_n)[/itex] is a sequence and x_n→0 then because sin(1/x_n) is bounded xsin(1/x)→0. So [itex]\displaystyle\lim_{x\rightarrow 0} ~~{xsin(1/x)}[/itex]=0

Attempt #2:
[itex]\displaystyle\lim_{x\rightarrow 0} ~~{xsin(1/x)}[/itex] = [itex]\displaystyle\lim_{x\rightarrow 0}~~ {\frac{(1/x)(x)(sin(1/x))}{(1/x)}}[/itex] = [itex](1/x)(x) ~~\displaystyle\lim_{x\rightarrow 0} ~~ {\frac{sin(1/x)}{(1/x)}}[/itex]= (1)(1) = 1

Can you guys tell me which is correct, and what is my error in the incorrect attempt? Thanks!

Answers and Replies

Would you happen to be familiar with what a taylor series is? Because if you know the series function for sine then the answer would be very clear,

nevertheless here is another way to look at it:

What happens to expression 1/X_n as X_n approaches zero...

Follow up to that what happens then to sine (1/Xn) as Xn approaches zero

Follow up to that what happens then to sine (1/Xn) as Xn approaches zero

as X_n→0, 1/X_n approaches ±∞. The limit as it approaches from the left does not equal the limit as it approaches from the right, so the limit of 1/X_n as X_n approaches 0 does not exist. So the limit of sin(1/x_n) also doesn't exist?

Can you pinpoint exactly where I went wrong in my attempts or is this question simply a special case?

don't think it does not exist JUST YET...

What is the range of sin(x)?

What happens to the actual graph as sin(1/x) approaches zero from -1 to 0 and from 1 to zero?

The limit as x approaches zero of x * sin(1/x)

is taking the limit as x approaches zero and multiplying it with the limit as sin(1/x) approaches zero...

So we're trying to find out what happens to the behavior as it gets closer to zero...

Keep in mind that sin of anything is restricted to a range of [-1, 1]

One additional clue, The function sin(1/x) oscillates increasingly faster as x approaches zero so at zero itself its undefined because it could be any number in [-1, 1]...

But what is the value of x as it approaches zero? and what happens if you multiply the value of sin(1/x) with x? Does it converge back to an answer from being undefined or is it still undefined?

Lastly:

Recalculate the Limit as x approaches 0 for sin(1/x)/(1/x) and tell me what answer you get

But what is the value of x as it approaches zero? and what happens if you multiply the value of sin(1/x) with x? Does it converge back to an answer from being undefined or is it still undefined?

Lastly:

Recalculate the Limit as x approaches 0 for sin(1/x)/(1/x) and tell me what answer you get

The range of sin x is [-1,1], so the range of sin (1/x) is also [-1,1]. Because the limit of x as x→0 = 0, multiplying this by sin(1/x) will give us 0 (because range of sin(1/x) is bounded). So I would think that the limit of (x)(sin1/x) as x→0 would equal 0.

So the limit of (sin1/x)/(1/x) is not defined, because sin (1/x) stays between [-1,1] but (1/x) will increase/decrease to ±∞ as x approaches 0. Because the bottom is increasing/decreasing to ±∞, it the limit of (sin 1/x)/(1/x) is also 0. Is this now correct?

I'm curious as to why I can't use the property that the limit of (sin x)/x as x→0 = 1 for this example, plugging in (1/x) for x?

The range of sin x is [-1,1], so the range of sin (1/x) is also [-1,1]. Because the limit of x as x→0 = 0, multiplying this by sin(1/x) will give us 0 (because range of sin(1/x) is bounded). So I would think that the limit of (x)(sin1/x) as x→0 would equal 0.

Use the sandwich or squeeze theorem.

So the limit of (sin1/x)/(1/x) is not defined, because sin (1/x) stays between [-1,1] but (1/x) will increase/decrease to ±∞ as x approaches 0. Because the bottom is increasing/decreasing to ±∞, it the limit of (sin 1/x)/(1/x) is also 0. Is this now correct?

First you say the limit isn't defined, and then you say it's also 0. Which is it?

I'm curious as to why I can't use the property that the limit of (sin x)/x as x→0 = 1 for this example, plugging in (1/x) for x?

It's because 1/x goes to ∞ as x goes to 0, so you have
[tex]\lim_{x \to 0} \frac{\sin (1/x)}{1/x} = \lim_{u \to \infty} \frac{\sin u}{u}[/tex]The second limit would equal 1 if u→0, but that's not what you have.

Use the sandwich or squeeze theorem.

It's because 1/x goes to ∞ as x goes to 0, so you have
[tex]\lim_{x \to 0} \frac{\sin (1/x)}{1/x} = \lim_{u \to \infty} \frac{\sin u}{u}[/tex]The second limit would equal 1 if u→0, but that's not what you have.

Could I also use the domination principle? Because sin(1/x) is between [-1,1] then |x*sin(1/x)| < |x| where x→0 so |x|→0 and so |x||sin(1/x)| also converges to 0?

Thank you for the explanation!!

Homework Statement

[itex]\displaystyle\lim_{x\rightarrow 0}~~ {xsin(1/x)}[/itex]

The Attempt at a Solution

I've attempted to solve this limit two different ways and get different answers.

Attempt #1:
If [itex](x_n)[/itex] is a sequence and x_n→0 then because sin(1/x_n) is bounded xsin(1/x)→0. So [itex]\displaystyle\lim_{x\rightarrow 0} ~~{xsin(1/x)}[/itex]=0

Attempt #2:
[itex]\displaystyle\lim_{x\rightarrow 0} ~~{xsin(1/x)}[/itex] = [itex]\displaystyle\lim_{x\rightarrow 0}~~ {\frac{(1/x)(x)(sin(1/x))}{(1/x)}}[/itex] = [itex](1/x)(x) ~~\displaystyle\lim_{x\rightarrow 0} ~~ {\frac{sin(1/x)}{(1/x)}}[/itex]= (1)(1) = 1

Can you guys tell me which is correct, and what is my error in the incorrect attempt? Thanks!

Another mistake in your second attempt is that you can't simply pull the factors containing x out of the limit since you're taking the limit as x→0. You can only pull constant factors out.

As vela has already suggested, use the squeeze theorem.

-|x| ≤ x sin(1/x) ≤ |x|

Could I also use the domination principle? Because sin(1/x) is between [-1,1] then |x*sin(1/x)| < |x| where x→0 so |x|→0 and so |x||sin(1/x)| also converges to 0?

This is essentially an application of the squeeze theorem. You have [itex]0 \le |x||\sin(1/x)| \le |x|[/itex] (note it should be ≤, not <, because sin(1/x) can equal 1), and because both the left and right go to 0, the middle must also go to 0.

So you've shown that
[tex]\lim_{x \to 0}\ |x\sin(1/x)| = 0[/tex]You still need to show why this implies
[tex]\lim_{x \to 0}\ x\sin(1/x) = 0[/tex]

So you've shown that
[tex]\lim_{x \to 0}\ |x\sin(1/x)| = 0[/tex]You still need to show why this implies
[tex]\lim_{x \to 0}\ x\sin(1/x) = 0[/tex]

Is this simply because |x| = 0 then x=0? So [itex]\lim_{x \to 0}\ x\sin(1/x)[/itex] must equal 0 because its absolute value limit is 0?

[tex]\lim_{x\to 0}{\left|x \sin{\frac{1}{x}}\right|}=0[/tex]
Or
[tex]\forall \varepsilon>0\hskip10pt \exists \delta>0 \hskip10pt\ 0<|x-0|<\delta \hskip10pt\ \left|\left|x\sin{\frac{1}{x}}\right|-0\right|<\varepsilon[/tex]
But, [tex]\left|x\sin{\frac{1}{x}}\right|=\left|\left|x\sin{\frac{1}{x}}\right|\right|<\varepsilon[/tex]
so we can say
[tex]\lim_{x\to 0}{x\sin{\frac{1}{x}}}=0[/tex]
Is this OK?

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