I was studying the limit of a multivariable function. I was given this example: Does $\lim_{(x, y) \to (0, 0)} f(x, y)$ exist, if $f(x, y) = \frac{xy^2}{x^2 + y^4}$?
The solution says if I approach the limit from $y = mx$, where $m$ is the gradient of a linear line, I get $$f(x, y) = f(x, mx) = \frac{x(mx)^2}{x^2 + (mx)^4} = \frac{m^2x}{1 + m^4x^2}$$
So $f(x, y) \rightarrow 0$ as $(x, y) \rightarrow (0, 0)$ along $y = mx$. This includes all the linear lines except for $x = 0$ as $x = 0$ represents the $y$-axis, where $m$ doesn't exist.
Now, if I let $x = 0$ and approach the limit along the $y$-axis, I get $$\lim_{y \to 0} f(0, y) = \lim_{y \to 0} \frac{0}{y^4} = 0$$
Therefore, I have showed that $\lim_{(x, y) \to (0, 0)} f(x, y) = 0$ along all the linear lines in the $xy$-plane, which really tempts me to conclude that $\lim_{(x, y) \to (0, 0)} f(x, y)$ exists.
However, the solution says that if I approach the limit along $x = y^2$, I get $$f(x, y) = f(y^2, y) = \frac{y^2 \cdot y^2}{(y^2)^2 + y^4} = \frac{y^4}{2y^4} = \frac{1}{2} \not= 0$$
Therefore, $\lim_{(x, y) \to (0, 0)} f(x, y)$ doesn't exist.
This confuses me. Because I thought that when I approached a limit along a curve (e.g. $x = y^2$), if I zoomed in enough, according to linear approximation, I could view the curve as a straight line and I was effectively approaching the limit along a straight line. In other words, no matter if I am approaching the limit along a straight line or along a curve, they should give me the same result, shouldn't they? Can anyone explain what is wrong with this thought? Thank you!
Update on 17 Nov 2023: I am still thinking about this question. And my question can be: I think approaching a point along all straight lines towards that point is enough to cover all directions. But apparently, it's not. Why not? Because near the point the curve is essentially straight if I zoom in a lot.