Is this method to check the continuity of multivariable functions correct?

Question

Consider $$f(x,y) =\begin{cases} \frac{\cos y\sin x}{x} & x\neq 0 \\ \cos y & x=0 . \end{cases}$$

Is $f$ continuous at $(0,0)$?

Method 1
$f(0,0) = 1.$
Also, $\lim_{(x,y)\to(0,0)} f({x,y})=1 = f(0,0).$ (Easy to calculate.)
Therefore, $f$ is continuous at $(0,0)$.

The problem I find with this method is that it didn't verify whether the limit exists or not at $(0,0)$. I did check $y=mx$ path to the point $(0,0)$, and it did give $1$ as the value of the limit. But obviously, I cannot verify all the existing paths to $(0,0)$, so the only way to check the existence of the limit is (I think) to use the epsilon-delta definition of limit. But if I have to use the epsilon-delta definition of limit to check the existence of the given limit, then I could very well use the epsilon-delta definition of continuity to check whether the given function is continuous at $(0,0)$. Then there is no need to find the limit or check its existence. If that's the case, then isn't Method 1 incomplete and in a way redundant as well?

Answer 1

HINT:

The idea is to start from the $\epsilon-\delta$ definition of continuity. So you write $$f(x+\delta_1,y+\delta_2)-f(x,y)=[f(x+\delta_1,y+\delta_2)-f(x,y+\delta_2)]+[f(x,y+\delta_2)-f(x,y)]$$

Answer 2

I think that it is much easier to do it as follows. Consider the function$$\begin{array}{rccc}\varphi\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}\frac{\sin x}x&\text{ if }x\ne0\\1&\text{ if }x=0.\end{cases}\end{array}$$Then $\varphi$ is continuous. It is clear that it is continuous at every point other than $0$, and it is also continuous at $0$, since $\lim_{x\to0}\frac{\sin x}x=1$. But then $f(x,y)=\varphi(x)\cos(y)$. Then $f$ is continuous, since it is the product of two continuous functions.

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If you really want to use the $\varepsilon-\delta$ definition, it can be used indeed. Note that, if $x\ne0$,$$f(x,y)-1=\frac{\sin x}x-1+\cos(y)-1-\left(\frac{\sin x}x-1\right)\bigl(\cos(y)-1\bigr).\tag1$$Given $\varepsilon>0$, take $\delta>0$ such that$$|x|<\delta\implies\left|\frac{\sin x}x-1\right|<\min\left\{\frac\varepsilon3,\sqrt{\frac\varepsilon3}\right\}\tag2$$and that$$|x|<\delta\implies\bigl|\cos(y)-1\bigr|<\min\left\{\frac\varepsilon3,\sqrt{\frac\varepsilon3}\right\}.\tag3$$So, if $\|(x,y)\|<\delta$, then $|x|,|y|<\delta$ and it follows nor from $(1)$, $(2)$, and $(3)$ that$$\left|f(x,y)-1\right|<\frac\varepsilon3+\frac\varepsilon3+\sqrt{\frac\varepsilon3}^2=\varepsilon.$$And it is clear that, if $x=0$, it is still true that $|f(x,y)-1|<\delta$.