Why is the limit along the y-axis 0?

For a math course, my course book computes the limit of the function $$f(x,y) = \dfrac{xy^2}{x^2+y^4}$$ at $$(0,0)$$ along the $$y$$-axis (along $$\mathbf{r}(t) = (0,t)$$). It finds $$\lim _{t \rightarrow 0}f(\mathbf{r}(t)) = \lim _{t \rightarrow 0}\dfrac{0}{0+t^4} = 0\\$$.

My question is why is this limit $$0$$ when plugging $$t=0$$ in will give $$\dfrac{0}{0}$$. I have tried to find similar problems online to compare but every time I find a similar problem it provides the answer $$0$$ without a good explanation.

• You're right, and this means it dose not have a limit at $(0,0)$. – Oolong milk tea May 11 at 5:32
• note that $f(0,y)=\frac{0\cdot y^2}{0^2+y^4}=0, y\ne 0$. – farruhota May 11 at 9:14

We are not plugging in values just yet. First we simplify, as $$t\neq0$$: $$\lim_{t\to 0}\frac0{0+t^4}=\lim_{t\to0}0$$ Now we can plug in $$0$$ for $$t$$ in the right-hand side and evaluate the limit. We see that it is $$0$$.
• You should mention that $0/t^4=0$ only for $t\neq 0$. – JustAnotherStackUser May 11 at 6:36
• @MaximillianJanisch $t\neq0$ is implicit when I write $\lim_{t\to0}$ in front of an expression. But yes, I could point it out explicitly. – Arthur May 11 at 7:27