Suppose I have a function $f:R^2 \to R$, where $R$ are the reals. I want to prove that $$\lim_{(x,y) \to (0,0)} f(x,y) = L \implies \lim_{x \to 0} f(x,mx^2) = L$$ by using a $(\varepsilon, \delta)$ proof.
My initial attempt was as so: We want to show that $\forall \varepsilon, \exists \delta$ such that $0<|x|<\delta \implies |f(x,mx^2) - L| < \varepsilon$.
We start from the assumption that $\lim_{(x,y) \to (0,0)} f(x,y)=L$, meaning that $$\forall \varepsilon, \exists \delta_1 \ s.t. \ 0<|(x,y)|<\delta_1 \implies |f(x,y) - L| < \varepsilon$$ I fix $m \in R$ and set $y = mx^2$, meaning I approach the limit point by the parabola. Thus, $$|(x,y)| = |(x,mx^2)| = |x|\sqrt{1+m^2x^2} < \delta_1$$ My initial goal was to somehow obtain $|x|$ by itself on 1 side of the equation, and obtaining an inequality kind of like $$|x| < g(m) \delta_1$$ where $g(m)$ is just some random function of $m$ and then set $\delta = g(m) \delta_1$, which would have completed the proof. However, it didn't take me long to realize that I'm unable to simplify $|(x,mx^2)| = |x| \sqrt{1+m^2x^2}$ in such a way that would leave me with an inequality with $|x|$ alone on one side.
I went with this approach as it previously worked out to prove that $$\lim_{(x,y) \to (0,0)} f(x,y) = L \implies \lim_{x \to 0} f(x,mx) = L$$ but I guess squares are just jerks. I'd really appreciate it if anybody could enlighten me or propose an alternative route. Honestly, I'm not sure how correct the theorem is, I'm just following the lecture notes of an introductory multivariable calculus course I'll be taking this year (meaning my tools are limited as far as math goes).