Find limit of multi variable function

[ananonanunes]

Homework Statement

Given the continuous function $g\colon \mathbb{R} \rightarrow \mathbb{R}$ so that $g(1)=7$, consider the function defined in $\mathbb{R}^{2}_{\backslash(1,0)}$ by$$f(x,y)=\frac {g(x)(x-1)^2y}{2(x-1)^4+y^2}.$$

Say whether the limit $\lim_ {(x,y) \rightarrow (1,0)} {f(x,y)}$ exists.

Relevant Equations

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This is what I did: $$\lim_ {(x,y) \rightarrow (1,0)} {\frac {g(x)(x-1)^2y}{2(x-1)^4+y^2}}=\lim_ {(x,y) \rightarrow (1,0)} {g(x)y\frac {(x-1)^2}{2(x-1)^4+y^2}}$$ I know that $\lim_ {(x,y) \rightarrow (1,0)} {g(x)y}=0$ and that $\frac {(x-1)^2}{2(x-1)^4+y^2}$ is limited because $0\leq (x-1)^2\leq 2(x-1)^4\leq 2(x-1)^4+y^2\Rightarrow 0 \leq\frac{(x-1)^2}{2(x-1)^4+y^2}\leq1$, so I concluded that the limit exists and it is 0.
I know this is wrong and I can understand that the limit cannot exist because if you calculate the limit for $y=(x-1)^2$ you get a solution different than 0. But I thought that having a function whose limit is 0 times a function that is limited was enough to conclude that the limit of their multiplication was 0. Is this wrong?

 

[BvU]

I did not know that
$$(x-1)^2\leq 2(x-1)^4$$
did I miss something ?

 

[ananonanunes]

I did not know that
$$(x-1)^2\leq 2(x-1)^4$$
did I miss something ?

You're right! I totally forgot about the numbers between 0 and 1. Thanks for the help

 

[pasmith]

Set $x = 1 + r\cos \theta$, $y = r \sin \theta$ and consider the limit $r \to 0$. Does it depend on $\theta$?

EDIT This is not sufficient to conclude that the limit exists, as the example I give below shows.

Instead consider what happens if you approach (1,0) along the path $(1 + t, At^2)$ for arbitrary $A$. If the limit exists, then the result should be independent of $A$.

 

[ananonanunes]

Set $x = 1 + r\cos \theta$, $y = r \sin \theta$ and consider the limit $r \to 0$. Does it depend on $\theta$?

It does. So if I have a two variable function and I want to calculate the limit for (x,y) to (0,0) I can just replace x and y with the polar coordinates and check whether it depends on \theta and if it does, there is no limit?

 

[WWGD]

Maybe I shouldn't, but I'd suggest you consider 2 or more random paths of approach. If you get the same limit on both, the limit likely exists. Good if you're stuck in an exam with little time left. Easy choices: along $y=x$ along $(0,y)$or $(x,0)$

 

[SammyS]

It does. So if I have a two variable function and I want to calculate the limit for (x,y) to (0,0) I can just replace x and y with the polar coordinates and check whether it depends on \theta and if it does, there is no limit?

Your missing one small detail given by @pasmith ; take the limit as $r\to 0$. Then you see if that limit depends on $\theta$, ...

 

[pasmith]

Maybe I shouldn't, but I'd suggest you consider 2 or more random paths of approach. If you get the same limit on both, the limit likely exists. Good if you're stuck in an exam with little time left. Easy choices: along $y=x$ along $(0,y)$or $(x,0)$

Consider $$f(x,y) = \frac{xy}{x^2 + y}.$$ Here approaching the origin along a straight line path we find $$\lim_{t \to 0} f(t\cos \theta,t\sin\theta) = \lim_{t\to 0} \frac{t\sin 2\theta}{2(t\cos^2 \theta + \sin \theta)} = 0.$$ However, we also find that for $C \neq 0$ we have $$\lim_{t \to 0} f\left(t, \frac{Ct^2}{t-C}\right) = C.$$ Hence $\lim_{(x,y)\to(0,0)} f(x,y)$ does not exist: there are points arbitrarily close to the origin where $f(x,y) = C$ for any $C$.

 

[WWGD]

@pasmith , not saying your method doesn't work, just offering a different approach, perspective.

 

[pasmith]

@pasmith , not saying your method doesn't work, just offering a different approach, perspective.

My point is that "consider 2 or more random paths of approach. If you get the same limit on both, the limit likely exists" is not accurate.

 

[WWGD]

My point is that "consider 2 or more random paths of approach. If you get the same limit on both, the limit likely exists" is not accurate.

" Likely"
It's a rule of thumb. Besides, the limits are different for $C\neq 0$, which doesn't qualify for my rule ( of thumb)