Why can't you break the limit of x*sin(1/x) into two separate limits?

[krackers]

I was wondering why when solving this limit, you are not allowed to do this:

{ lim }_{ x\rightarrow 0 }\quad xsin(\frac { 1 }{ x } )

Break this limit into:

{ lim }_{ x\rightarrow 0 }\quad x\quad *\quad { lim }_{ x\rightarrow 0 }\quad sin(\frac { 1 }{ x } )

Then, since, sin(1/x) is bounded between -1 and 1, and lim x-> 0 (x) is 0, the answer should be 0.

 

[lugita15]

I was wondering why when solving this limit, you are not allowed to do this:

{ lim }_{ x\rightarrow 0 }\quad xsin(\frac { 1 }{ x } )

Break this limit into:

{ lim }_{ x\rightarrow 0 }\quad x\quad *\quad { lim }_{ x\rightarrow 0 }\quad sin(\frac { 1 }{ x } )

Then, since, sin(1/x) is bounded between -1 and 1, and lim x-> 0 (x) is 0, the answer should be 0.

That's not rigorous enough, because lim_{x\rightarrow 0}sin\frac{1}{x} doesn't exist. But what you can do is say that -x\leq xsin\frac{1}{x}\leq x for all x\neq 0, and lim_{x\rightarrow 0}(-x)=lim_{x\rightarrow 0}(x)=0, so by the squeeze theorem lim_{x\rightarrow 0}xsin\frac{1}{x}=0

 

[marecrisium]

"That's not rigorous enough, because lim x→0 sin1 x doesn't exist."

I have to agree with this person. That limit doesn't really make sense. It only loops through a ratio. Because it loops through a ratio along with another variable, it makes the limit of the function an oscillating one, you have to check the derivatives instead.

1*sin(1/x) - x(-1/x^2)(sin1/x)(cos1/x)
take apart the trigonometric functions... because they only loop through -1 and 1
1 + (1/x)
... and you get an idea of where the limit is going as approaches 0
This is showing you that the derivative is unstable. It just does not level off.

lim(x->0) for sin(1/x) will just continue to oscillate as sin(1/x) flucuates through ratios. This should support the person's above statement that the limit doesn't really exist, and that values will always be between 1 and -1.

I am not so sure however myself. I haven't done some of this kind of work in a year. Can anyone add more input? I can't remember the exact procedure for checking this one's limit.

The actual limit cannot be zero as a result of the oscillating functions. Just think, if 1/x allowed x to get smaller and smaller, 1/x as a function would continue to grow. It ends up as sin (+t) where t continues to increase without bounds. The sine function still can only go to one. Even when being subjected to proportion adjustments from it's neighbor variable x. Hence, X * sin(+t) cannot have a zero limit, it really looks more like (k)sin(+t) if you think about it. It will never leave the 1*k and -1*k range.

It is hard to explain, but if you check your calculator it might make a little more sense. Just use .01 sized values for x in the function sin(1/x) Notice that these values when multiplied with the same value of nearby x will still be in flux. Using my own calculator, I see now when checking with .01 sized values that 0 in fact cannot be the limit, check for yourself!

Punch this is in and say I showed you how...

xsin(1/x) and then use DELTATBL=.001

check the tables and you will see exactly what I mean...

It is from here that you should be able to say that the limit might actually be zero...

 

[DonAntonio]

I was wondering why when solving this limit, you are not allowed to do this:

{ lim }_{ x\rightarrow 0 }\quad xsin(\frac { 1 }{ x } )

Break this limit into:

{ lim }_{ x\rightarrow 0 }\quad x\quad *\quad { lim }_{ x\rightarrow 0 }\quad sin(\frac { 1 }{ x } )

Then, since, sin(1/x) is bounded between -1 and 1, and lim x-> 0 (x) is 0, the answer should be 0.

There's no mathematical sound meaning to \,\lim f(x)g(x)=\lim f(x)\cdot \lim g(x)\, if any of these limits doesn't exist, yet

you too can tell what you said, namely: if
\lim_{x\to\ x_0}f(x)=0\,\,\text{and}\,\,\,g(x)\,\,\,\text{ is bounded in some neighbourhood of}\,\,x_0
then \lim_{x\to x_0}f(x)g(x)=0

The \,\epsilon - \delta is very simple.

DonAntonio

 

[who_]

It isn't too hard if you think about it. Consider the sequence {x_n} = {1/pi, 1/2pi, 1/3pi, ...}.
Then sin(1/x_n) = 0 and thus goes to 0 as n goes to infinity.

Similarly, consider {y_n} = {2/pi, 2/5pi, 2/9pi, ...}. Then sin(1/y_n) = 1 and thus goes to 0 as n goes to infinity.

Since you have two subsequences approaching 0 such that the original function approaches two different vales, the limit cannot exist.

 

[who_]

It isn't too hard if you think about it. Consider the sequence {x_n} = {1/pi, 1/2pi, 1/3pi, ...}.
Then sin(1/x_n) = 0 and thus goes to 0 as n goes to infinity.

Similarly, consider {y_n} = {2/pi, 2/5pi, 2/9pi, ...}. Then sin(1/y_n) = 1 and thus goes to 0 as n goes to infinity.

Since you have two subsequences approaching 0 such that the original function approaches two different vales, the limit cannot exist.

 

[DonAntonio]

It isn't too hard if you think about it. Consider the sequence {x_n} = {1/pi, 1/2pi, 1/3pi, ...}.
Then sin(1/x_n) = 0 and thus goes to 0 as n goes to infinity.

Similarly, consider {y_n} = {2/pi, 2/5pi, 2/9pi, ...}. Then sin(1/y_n) = 1 and thus goes to 0 as n goes to infinity.

Since you have two subsequences approaching 0 such that the original function approaches two different vales, the limit cannot exist.

The wanted limit is when x\to 0 , not to infinity.

DonAntonio

 

[who_]

The wanted limit is when x\to 0 , not to infinity.

DonAntonio

Yes, x_n and y_n are going to 0. The index n is going to infinity. I merely created sequences x_n and y_n that approach 0 such that sin(1/x_n) and sin(1/y_n) approach different values.

 

[Millennial]

Substitute x = 1/t. Then the limit we want becomes \displaystyle \lim_{t\to\infty}\frac{\sin(t)}{t}. And, sine is a bounded function: -1 \leq \sin(x) \leq 1. This puts the limit in the form of the direct application of the squeeze theorem.

Can you take it from there?

 

[DonAntonio]

Yes, x_n and y_n are going to 0. The index n is going to infinity. I merely created sequences x_n and y_n that approach 0 such that sin(1/x_n) and sin(1/y_n) approach different values.

It doesn't matter what the sine approaches: it is bounded everywhere and, in particular, it is bounded in any neighbourhoodo f zero. Since the other functions approaches zero the whole thing approaches zero, too.

DonAnronio

 

[who_]

It doesn't matter what the sine approaches: it is bounded everywhere and, in particular, it is bounded in any neighbourhoodo f zero. Since the other functions approaches zero the whole thing approaches zero, too.

DonAnronio

I know what the OP's original question is. The point of my post was to help illustrate why sin(1/x) does not have a limit as x goes to 0.

 

[DonAntonio]

I know what the OP's original question is. The point of my post was to help illustrate why sin(1/x) does not have a limit as x goes to 0.

Ok. Perhaps next time it'd a good idea to quote from the OP what you're addressing. In your first post you talked

about "the original function", which goes to zero, and apparently you were reffering ONLY to part

of it, namely to \,\sin 1/x\, . This may be confusing.

DonAntonio.

 

[Mandlebra]

Ok. Perhaps next time it'd a good idea to quote from the OP what you're addressing. In your first post you talked

about "the original function", which goes to zero, and apparently you were reffering ONLY to part

of it, namely to \,\sin 1/x\, . This may be confusing.

DonAntonio.

I thought it was easily inferred...