Find $$\lim_{(x,y) \to (0,0)}\frac{3x^{2}y}{x^{2}+y^{2}}$$ if it exists.
From my textbook, it said that the limits along the parabolas $$y=x^{2}\text{ and } x=y^{2}$$ also turn out to be 0.
I couldn't figure out why, as if we set $$x=y^{2}$$, for example
then $$\lim_{(x,y) \to (0,0)}\frac{3x^{4}}{x^{2}+(x^{2})^{2}}$$
$$\lim_{(x,y) \to (0,0)}\frac{3x^{4}}{x^{2}+x^{4}}$$ $$\lim_{(x,y) \to (0,0)}\frac{3}{0+1}$$ then the answer is supposed to be $3$? right?
Note that
$$\lim_{x\to \infty}\frac{3x^{4}}{x^{2}+x^{4}}=3$$ but $$\lim_{x\to 0}\frac{3x^{4}}{x^{2}+x^{4}}=\lim_{x\to 0}\frac{3x^{2}}{1+x^{2}}=0.$$
On $y=x^2$ we have $$ \frac{3x^2y}{x^2+y^2}=\frac{3x^4}{x^2+x^4}= \frac{3x^4}{x^2(1+x^2)}=\frac{3x^2}{1+x^2} $$ which has limit $0$ when $x\to0$.
On $x=y^2$ we have $$ \frac{3x^2y}{x^2+y^2}=\frac{3y^5}{y^4+y^2}=\frac{3y^5}{y^2(y^2+1)} =\frac{3y^3}{y^2+1} $$ which again has limit $0$ for $y\to0$.
A smaller (in magnitude) denominator means a larger number in magnitude (given numerator are the same).
$$|\frac{3x^2y}{x^2+y^2}| \leq |\frac{3x^2y}{x^2}|=|3y| \to 0$$
Then conclude with squeeze theorem.
Your error is that $\frac{x^2}{x^4}=\frac{1}{x^2} \to \infty$ not zero.
So the denominator goes to infinity and hence your limit goes to $0$ as claimed in the book.