Why doesn't the limit exist for this function at (0,0)?

[NanoMath]

Homework Statement

I have to show that the following function does not have a limit as (x,y) approaches (0,0)

The Attempt at a Solution

I tried taking different paths for example y=x or y=0 and switching to polar coordinates, but I don't get anywhere.

 

[pasmith]

Consider approaching the origin along the line x = 0.

 

[NanoMath]

Right so the first term cancels and I am left with: y sin ( 1/ y² ) + sin ( 1 / y² ) .
So what do I conclude by taking the limit of this?
y sin ( 1/ y² ) vanishes since y goes to 0. I am a bit confused on what to do with sin ( 1/y²).
Can I claim that limit doesn't exist because sin ( 1/ y² ) alternates as y goes to zero and therefore doesn't have unique answer ?

 

[pasmith]

Right so the first term cancels and I am left with: y sin ( 1/ y² ) + sin ( 1 / y² ) .
So what do I conclude by taking the limit of this?
y sin ( 1/ y² ) vanishes since y goes to 0. I am a bit confused on what to do with sin ( 1/y²).
Can I claim that limit doesn't exist because sin ( 1/ y² ) alternates as y goes to zero and therefore doesn't have unique answer ?

How would you justify that rigorously?

 

[STEMucator]

You can explain the dense oscillations near the origin by recalling $\text{sin}(y) = 0$ for $y = n \pi, n \in \mathbb{Z}$.

So $\text{sin} \left( \frac{1}{y} \right) = 0$ for $y = \frac{1}{n \pi}, n \in \mathbb{Z}, n \neq 0$.

Therefore we can say $\text{sin} \left(\frac{1}{y^2} \right) = 0$ for $y = \frac{1}{\sqrt{n \pi}}, n \in \mathbb{Z}, n \neq 0$.

There is a dense population of these points near zero; think about the interval $(0, \frac{1}{\sqrt{\pi}}]$.

As for showing it formally, you're going to have to squeeze some effort out, and you might need a sandwich to have enough energy.

 

[NanoMath]

Would it be okay to take sequence a_n = √2/√nπ . Then this sequence obviously converges to zero as n goes to infinity. But f(a_n) alternates between 1,0,-1,0,...

 

[STEMucator]

Would it be okay to take sequence a_n = √2/√nπ . Then this sequence obviously converges to zero as n goes to infinity. But f(a_n) alternates between 1,0,-1,0,...

There is infinitely many points approaching zero:

$$\frac{1}{\sqrt{\pi}}, \frac{1}{\sqrt{2 \pi}}, \frac{1}{\sqrt{3\pi}}, \frac{1}{\sqrt{4\pi}}, \frac{1}{\sqrt{5 \pi}}, \frac{1}{\sqrt{6 \pi}}, ...$$

Group these into triplets:

$$\left( \frac{1}{\sqrt{\pi}}, \frac{1}{\sqrt{2 \pi}}, \frac{1}{\sqrt{3\pi}} \right), \left( \frac{1}{\sqrt{4\pi}}, \frac{1}{\sqrt{5 \pi}}, \frac{1}{\sqrt{6 \pi}} \right), ...$$

Notice for each triplet of points inside $(0, \frac{1}{\sqrt{\pi}}]$, the function $\text{sin} \left(\frac{1}{y^2} \right)$ will oscillate from $0$ to $-1$, then to $0$, then to $+1$ and then back to $0$.

The triplets need not be ordered as I've shown. You can group different triplets and the function will oscillate in the same fashion, but potentially in a different order. For example, using:

$$\left( \frac{1}{\sqrt{2 \pi}}, \frac{1}{\sqrt{3\pi}}, \frac{1}{\sqrt{4\pi}} \right)$$

Will produce a different sequence for the oscillation (0,1,0,-1,0), but the behavior will be the same for each triplet. Namely, $\left| \text{sin} \left(\frac{1}{y^2} \right) \right| \leq 1$.

 

[momoko]

Simply speaking, we choose epsilon=0.9, we cannot get a corresponding delta to satisfy the epsilon-delta definition of limit.