I need to determine if the limit as $\mathbf{x}\rightarrow \mathbf{0}$ exists for the following functions: ($f:\mathbb{R}^2 - \{(0,0^T)\} \rightarrow \mathbb{R}$)
- $f(x_1,x_2) = \frac{x_1x_2}{\sqrt{x_1^2+x_2^2}}$
- $f(x_1,x_2)=\frac{x_2^2-x_1^2}{x_1^2+x_2^2}$
- $f(x_1,x_2)=\frac{x_1 x_2^2}{x_1^2+x_2^4}$
The solution given is:
- The limit exists and is equal to $0$, by definition of limit at a point.
- The limit does not exist, by considering the trajectory $(t,\alpha t)^T$, where $\alpha$ is some constant parameter.
- The limit does not exist, by considering the trajectory $(\alpha t^2,t)^T$, where $\alpha$ is some constant parameter.
The problem I have is:
In the case of a single variable function, I can sketch the graph of the given function around the point of interest to see if the limit exists or does not exists, then structure my workings accordingly.
But in the two variable case, I am unable to sketch the contour plot (calculators and graph calculators are not allowed) as I don't see how setting $f(x_1,x_2)=c$ where $c$ is some constant can be further simplified to some conic section, line, etc.
Therefore, given the above functions (and any two variable function in general) how do you determine that the limit exist or not exist at a given point before showing it formally?
In a large amount of cases, converting the function into polar coordinates works very well. For example, for $f(x_1,x_2) = \frac{x_1x_2}{\sqrt{x_1^2+x_2^2}}$ converts into $$f(r,\phi) = \frac{r^2\cos\phi\sin\phi}{r}$$
and the limit as $r\to 0$ is easy to calculate.