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.

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

  • $\begingroup$ You should mention that $0/t^4=0$ only for $t\neq 0$. $\endgroup$ – JustAnotherStackUser May 11 at 6:36
  • 2
    $\begingroup$ @MaximillianJanisch $t\neq0$ is implicit when I write $\lim_{t\to0}$ in front of an expression. But yes, I could point it out explicitly. $\endgroup$ – Arthur May 11 at 7:27

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