Limits in multivariable calculus

[Chris18]

1. Find if the limit exist: sin (x^3 + y^3) / (x + y)
(x,y)-> (0,0)
So I am starting solving this by using polar coordinates form and I get to lim= sin r^3 ( cos^3θ + sin^3θ) / r ( cosθ + sinθ) = lim r^2 ( cos^2Θ + sin^2Θ) My question is ok so far and how should I procceed from now on?

 

[andrewkirk]

An alternative (and, I suspect, easier) approach is to use the factorisation $x^3+y^3=(x+y)(x^2-xy+y^2)$ together with the small angle approximation $\lim_{\theta\to 0}\frac{\sin\theta}\theta=1$.

 

[Chris18]

That's a good solution indeed thanks a lot but still I am interested to see how my way will go on? If I have r^2( cos^2θ + sin^2θ) can I say that the () is zero or should I do something more? Thanks in advance

 

[Ray Vickson]

1. Find if the limit exist: sin (x^3 + y^3) / (x + y)
(x,y)-> (0,0)
So I am starting solving this by using polar coordinates form and I get to lim= sin r^3 ( cos^3θ + sin^3θ) / r ( cosθ + sinθ) = lim r^2 ( cos^2Θ + sin^2Θ) My question is ok so far and how should I procceed from now on?

No: $(\cos^3 \theta + \sin^3 \theta) / (\cos \theta + \sin \theta) \neq \cos^2 \theta + \sin^2 \theta$.

 

[I like Serena]

That's a good solution indeed thanks a lot but still I am interested to see how my way will go on? If I have r^2( cos^2θ + sin^2θ) can I say that the () is zero or should I do something more? Thanks in advance

As @Ray Vickson pointed out, your expression is not correct.
Instead we have:
$$\frac{\cos^3 \theta + \sin^3 \theta}{\cos \theta + \sin \theta} = \cos^2 \theta - \cos\theta\sin\theta + \sin^2 \theta = 1 - \frac 12 \sin\theta$$
which is effectively what @andrewkirk suggested.

It is slightly different from yours, but the following argument is the same.
We can see that whatever $\theta$ is, the absolute value of the expression is less than or equal to 3/2.
When we multiply it by $r^2$ the result will go to zero when r does, as does the sine that is taken subsequently.
Therefore the limit exists and is 0.

 

[Ray Vickson]

As @Ray Vickson pointed out, your expression is not correct.
Instead we have:
$$\frac{\cos^3 \theta + \sin^3 \theta}{\cos \theta + \sin \theta} = \cos^2 \theta - \cos\theta\sin\theta + \sin^2 \theta = 1 - \frac 12 \sin\theta$$
which is effectively what @andrewkirk suggested.

It is slightly different from yours, but the following argument is the same.
We can see that whatever $\theta$ is, the absolute value of the expression is less than or equal to 3/2.
When we multiply it by $r^2$ the result will go to zero when r does.
Therefore the limit exists and is 0.

Yes, indeed----almost. The function has the form 0/0 along the line $y = -x$. I don't know whether that invalidates the whole concept of a limit, since when we take a limit in 2 dimensions we are supposed to allow $(x,y) \to (0,0)$ and any direction or along any curve. The limit-point (0,0) itself is not a problem, even though it, too, gives 0/0. I am genuinely puzzled by this question.

 

[I like Serena]

Yes, indeed----almost. The function has the form 0/0 along the line $y = -x$. I don't know whether that invalidates the whole concept of a limit, since when we take a limit in 2 dimensions we are supposed to allow $(x,y) \to (0,0)$ and any direction or along any curve. The limit-point (0,0) itself is not a problem, even though it, too, gives 0/0. I am genuinely puzzled by this question.

Since those points are not part of the domain of the function, they do not participate in the evaluation of the limit do they?
We would only pick directions or curves that are inside the domain.