Question

Let exFormula1" title="f(x,y) = \frac{x^{2} y}{x^{4}+y^{2} }" alt="f(x,y) = \frac{x^{2} y}{x^{4}+y^{2} }" align="absmiddle" class="latex-formula"> .
Which of the following is true about $\lim_{(x,y) \to \((0,0)} f(x,y)$ ?

Answer 1

The correct answer is C.

The limit in A does exist:

$\displaystyle \lim_{x\to0} f(x,0) = \lim_{x\to0} \frac0{x^4} = 0$

The limit in B also exists: for any $k\in\Bbb R$,

$\displaystyle \lim_{x\to0} f(x,kx) = \lim_{x\to0} \frac{kx^3}{x^4+k^2x^2} = \lim_{x\to0}\frac{kx}{x^2+k^2} = 0$

But this alone does not prove the 2D limit exists. $y=kx$ only captures all the paths through the origin that are straight lines.

The limit in C also exists, but it's not the same as either of the limits along the paths used in A and B.

$\displaystyle \lim_{x\to0} f(x,x^2) = \lim_{x\to0} \frac{x^4}{2x^4} = \frac12$

That this value is non-zero tells us the original limit does not exist.

The claim in D is generally not correct. That $f(0,0)$ is undefined does not automatically mean the limit doesn't exist. A simpler example:

$\displaystyle \lim_{x\to0} \frac{x}{x} = \lim_{x\to0} 1 = 1$

yet $\frac00$ is undefined.