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Now I have learnt that to prove a function of 2 variables exists we must have the both the repeated limits as equal, which is $\lim_{(x=0,y\to0)}f(x,y) = \lim_{(x\to0,y=0)}f(x,y)$ , now in this case $f(x,y)= \frac{2y^2}{\sqrt{x^2+xy}}$ so $\lim_{(x=0,y\to0)}f(x,y) = \lim_{(x=0,y\to0)} 2y^2/\infty$ which is not defined ,while for $\lim_{(x\to0,y=0)}f(x,y) = 0$ which is I think is enough to prove that this function( $\frac{2y^2}{\sqrt{x^2+xy}}$) does not exist as $(x,y) \to (0,0)$

Yet my book says it does exist , and I really can't find here where I go wrong , so I decided to ask here.

(P S : I guess i would be ridiculous to add this but at point $(0,y)$ my function is undefined ! doesn't that mean I simply don't need the point $(x,0)$ at all to prove it exists because its undefined and that straight off finishes the matter)

Arnav Das
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3 Answers3

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The limit cannot exist, since there are arbitrarily large values of $f(x,y) =\frac{2y^2}{\sqrt{x^2+xy}}$ in any neighborhood of $(0,0)$.

To wit, consider the open ball of radius $\varepsilon$ centered on $(0,0)$. Then we have $$ \lim_{x\to 0^+} f(x,\varepsilon/2) = +\infty $$ and this limit all happens within the ball, so there are arbitrary large values of $f$ within the ball. So $f$ cannot have any finite limit.

And the limit cannot be $+\infty$ either because $$ \lim_{t\to 0} f(t,t) = \lim_{t\to 0} \frac{2t^2}{t} = 0 $$

There's also the minor matter that $f$ is not even defined in any punctured neighborhood of $(0,0)$ -- both because of the division by zero when $x=0$, and because for any positive $y$ and $x\in(-y,0)$, the quantity $x^2+xy=x(x+y)$ will be negative, so the square root doesn't exist.

hmakholm left over Monica
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That's tricky, and I guess it depends on the interpretation. Because at points $(0,y)$ the expression in your limit is not even defined.

But you can try for instance with $x=y^4$, $y>0$. Then, using that if $y<1$ then $y^8<y^5$ and $$ \frac{2y^2}{\sqrt{x^2+xy}}=\frac{2y^2}{\sqrt{y^8+y^5}}\geq \frac{2y^2}{\sqrt{y^5+y^5}}=\frac{\sqrt2}{\sqrt{|y|}} $$

Martin Argerami
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Choose the curve $y=x^3-x$ s.t. $$lim_{x\rightarrow 0}\frac{2(x^3-x)^2}{\sqrt{x^2+x(x^3-x)}}=lim_{x\rightarrow 0} 2(x^2-1)^2=2$$ hence limit is not unique.

Nitin Uniyal
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