# Lim x->0+ 1/x sin(1/x) = ?

mr_coffee
hello everyone, im' really confused on how this is going to work...
I can't even graph this function. Does anyone know what this would equal and why? I know i don't have any work, because I'm totally lost on how I'm suppose to figure this out, if i had a graph it would be easy. Thanks! or lim x->0+ x*sin(1/x); I know this is calc I stuff and I'm in calc III but I'm helping out a friend.

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Do a change of variables: Let u=1/x.

phreak
You must first simplify 1/x*sin(1/x), which would come out to [sin(1/x)]/x

Use L'Hopital's Rule to differentiate both sides. The derivative of [sin(1/x)] = cos(1/x)*(-1/x^2) and the derivative of x = 1, right?

So the problem simplifies to lim(x-->0) cos(1/x)*(-1/x^2).

Both cos(1/x) and (-1/x^2) are undefined at x = 0, so what's the answer? There is none. It doesn't exist.

If you graph this, you'll see that as x-->0 from the left side, y goes down to negative infinity, and as x-->0 from the right side, y goes up to positive infinity. And since both sides don't approach the same number, there is no limit. (And even if you approach it from different sides, you get a nonexistent answer.)

(Can someone confirm the above method where I used L'Hopital's? I think it's correct, but I'm uncertain.)

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phreak: At Physics Forums we avoid posting solutions until it becomes evident that the person asking the question has exhausted every attempt to solve the problem himself. Until then, we give nudges in the right direction.

phreak said:
You must first simplify 1/x*sin(1/x), which would come out to [sin(1/x)]/x

Use L'Hopital's Rule to differentiate both sides.

Hold up there: You can only use L'Hopital for the indeterminate forms $0/0$ and $\infty/\infty$. This is neither.

phreak
Oh, that's right. I was thinking of something else... Sorry.

mr_coffee
Thanks for the replies, but If i let u = 1/x, i get u*sin(u). I'm not suppose to integrate by substitution am i? its a limit problem, so phreaks method is incorrect?

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mr_coffee said:
but If i let u = 1/x, i get u*sin(u). I'm not suppose to integrate by substitution am i? Where do you see an integration sign?

No, you're just changing the variable. So you get u*sin(u). Now, as x approaches 0, what does u approach? Rewrite the limit using that change of variables, and evaluate it.

its a limit problem, so phreaks method is incorrect?

Yes, it's incorrect for the reason I stated.

mr_coffee
ohh sorry i don't know what i was thinking, i never knew you could let u = 1/x. well as x approaches 0 on the right side it goes to infinity and x approaches 0 on the left side its negative infitiy so there is no limit there. So does this prove that 1/x*sin(x) = no limit?

Homework Helper
mr_coffee said:
ohh sorry i don't know what i was thinking, i never knew you could let u = 1/x. well as x approaches 0 on the right side it goes to infinity and x approaches 0 on the left side its negative infitiy so there is no limit there. So does this prove that 1/x*sin(x) = no limit?
A good way to show that
$$\lim_{x\rightarrow 0}f(x)$$
does not exist would be to show that for some value h>0
any open interval contaning 0 will contain points x and y such that
|f(x)-f(y)|>h

mr_coffee
lurflurf, that sounds good but i don't understand it... |f(x)-f(y)| >h, how is that showing there is no limit?

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mr_coffee said:
lurflurf, that sounds good but i don't understand it... |f(x)-f(y)| >h, how is that showing there is no limit?
If the limit exist then if x is close enough to 0, f(x) and f(y) will also be close together as they are both close to the limit.
Thus if |f(x)-f(y)| cannot be made small by making |x-y| small, the limit cannot exist.
It is important to note |xy|>0.
Two nice example of this are (for x->0) 1/x and sin(1/x)
Consider the function
f(x)=x+10^-50 if 1/x=n^n for some integer n
f(x)=x otherwise
clearly the limit does not exist because
|2n^-n-n^-n| can be made very small by taking n (integral) large, but no matter how large n is
|f(2n^-n)-f(n^-n)|=10^-50>10^-57

Homework Helper
I do not know what limit you wish to show does not exist.
Many limits have been stated that are not the same.
The suggested change of variable is for the two variations
$$\lim_{x\rightarrow 0^+} \ x \ \sin\left(\frac{1}{x}\right)=\lim_{u\rightarrow\infty} \ \frac{\sin(u)}{u}$$
$$\lim_{x\rightarrow 0^+} \ \frac{1}{x} \ \sin\left(\frac{1}{x}\right)=\lim_{u\rightarrow\infty} u \ \sin(u)$$

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phreak said:
You must first simplify 1/x*sin(1/x), which would come out to [sin(1/x)]/x
That is not really a simplification
Use L'Hopital's Rule to differentiate both sides. The derivative of [sin(1/x)] = cos(1/x)*(-1/x^2) and the derivative of x = 1, right?
Both sides of what? Why use L'Hopital's rule at all?
So the problem simplifies to lim(x-->0) cos(1/x)*(-1/x^2).

Both cos(1/x) and (-1/x^2) are undefined at x = 0, so what's the answer? There is none. It doesn't exist.
The functions need not exist at zero as this is a limit. The limit also does not exist, but it is important to recall that L'Hopital is only valid when the L'Hopital limit exist. When the L'Hopital limit does not exist no conclusion about the original limit can be made.
If you graph this, you'll see that as x-->0 from the left side, y goes down to negative infinity, and as x-->0 from the right side, y goes up to positive infinity. And since both sides don't approach the same number, there is no limit. (And even if you approach it from different sides, you get a nonexistent answer.)

(Can someone confirm the above method where I used L'Hopital's? I think it's correct, but I'm uncertain.)
bold type inside quote is my comments

mr_coffee
thank u so much, that makes sense now!