Limit in multivariable-calculus Let $\ell$ be a straight line through origo. Determine the limit to the restriction of $$f(x,y)=xye^{-x^2y^2}$$ to $\ell$ when $x^2+y^2 \to \infty$. Also, investigate the limit $$\lim_{x^2+y^2 \to \infty} f(x,y). $$
How should I think about the restriction? As I understand it, we have a line $\ell(t) = kt, k \in \mathbb R$ and I want $f$ to follow this path which would give us $$f(t,kt) = \frac{tkt}{e^{t^2(kt)^2}} $$
and if either $t$ or $kt$ is fixed to its minimum value ($0$) we get, by letting the other variable approach $\infty$, 
$$\lim_{t^2 \to \infty} f(t,0) = \frac{t \cdot 0}{e^{t^2(0)^2}} =0 $$
and
$$\lim_{(kt)^2 \to \infty} f(0,kt) = \frac{0 \cdot kt}{e^{0^2(kt)^2}} = 0$$ 
and so the limit should be equal to 0. Is this correct? And how do I "realize" that the second limit doesn't exist?
 A: There's some misunderstanding. It is correct that you fixed a line $y = kx$ (that is, $k$ is the slope of the line), and then you let $x \to \infty$ (So by that equation, $y \to \infty$ also). 
Now you have 
$$f(x, kx) = \frac{xkx}{e^{x^2 k^2x^2}} = \frac{kx^2}{e^{k^2 x^4}}$$
So there are two case: 


*

*$k=0$: Then $f(x, 0)  =0$ for all $x$ and so $\lim_{x\to \infty} f(x, 0) = 0$.

*$k\neq 0$: The bottom is growing much faster than the top, so $\lim_{x\to \infty} f(x, kx) = 0$
As a result, when restricted to all $\ell$ of the form $\{y = kx\}$, we have 
$$(*)\ \ \ \lim_{x\to \infty} f(x, kx) = 0.$$
However, this (*) is not the same as saying that 
$$\lim_{x^2 +y^2 \to \infty} f(x, y) = 0$$
Indeed, this is not true: Consider another curve $y = \frac{1}{x}$. Then 
$$f(x, \frac{1}{x}) = e^{-1}$$
for all $x$. So $\lim_{x\to \infty} f(x, \frac{1}{x}) = e^{-1}\neq 0$. This shows that the limit 
$$\lim_{x^2 +y^2 \to \infty} f(x, y) $$
does not exist. 
A: Note that along $y=1/x,x>0, f(x,y) = e^{-1}.$ So as $x^2 + y^2 \to \infty $ along this curve, $f(x,y) \not \to 0.$ Since we know that $f \to 0$ on lines through the origin, this shows that $\lim_{x^2+y^2\to \infty}f(x,y)$ does not exist.
