Investigating continuity for function at given point Given the function
$$
f(x,y)=\begin{cases}\big|1+xy^2\big|^\dfrac{1}{x^2+y^2} & \quad\hfill (x,y)\neq(0,0)\\\\ 1 &\quad\hfill (x,y)=(0,0)
\end{cases}
$$
investigate whether the function is continuous at $(0,0)$.
Usually, I claim $p\in\mathbb{R}$ such that $y=px$ , $x\rightarrow 0$, placing them in $\displaystyle \lim_{x\rightarrow0}f(x,y)$ and seeing how that works out. If $\displaystyle \lim_{x\rightarrow0}f(x,y)=f(0,0) $ then (according to how I was taught) the function is continuous at $(0,0)$.
In this specific exercise, I can't seem to solve it using methods I know, i.e the one explained above, or just choosing $y$ to be any variation of $x$ (e.g. $y=\sqrt x $).
So now I'm at a standstill in my thoughts.
Any hints/tips would be really helpful!
Thanks!
 A: The trick you have is not the definition of continuity at $(0,0)$. By setting $y=px$ with various values of $p$ and considering $\lim_{x\to 0}f(x,px)$, if any these limits do not agree or do not exist, you can tell that $f$ is not continuous at $(0,0)$. But even if they all equal to $f(0,0)$, you can not conclude that $f$ is continuous at the origin. See these two related questions:
Two variable function that's continuous on all linear paths, but nevertheless discontinuous
Example of a function that is not continuous at $(0, 0)$ but continuous when restricted to any curve approaching the origin

When $(x,y)\ne (0,0)$, rewrite the function as
$$
f(x,y)=\exp\left(\frac{\ln|1+xy^2|}{x^2+y^2}\right)
$$
Now by continuity of the exponential function, you want to investigate whether
$$
\lim_{(x,y)\to(0,0)}\frac{\ln|1+xy^2|}{x^2+y^2}=0\;.
$$
By Taylor's expansion of the natural logarithm, near $(0,0)$, you can write
$$
\ln|1+xy^2|=xy^2+g(x,y)
$$
for some function $g$ where $|g(x,y)|\le C|(xy^2)^2|$. But $\lim_{(x,y)\to(0,0)}\frac{g(x,y)}{x^2+y^2}=0$, so it suffices to look at
$$
\lim_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^2}
=\lim_{(x,y)\to(0,0)}x\cdot \frac{1}{(x/y)^2+1}\;.
$$
But this is zero because
$$
|\frac{1}{(x/y)^2+1}|\le 1\;.
$$
A: Your approach is not correct, using paths is only allowed to prove that a limit does not exist or in some cases to improve our intuition as to where the function approaches around a given point, nothing more.
Now, by definition

*

*$f(0,0)=1$
Also, setting the change of variables to polar coordinates $$\begin{cases}x=r\cos \theta,\\y=r\sin \theta\end{cases}$$ with $r\in \mathbb{R}^{+*}$ and $\theta\in [0,2\pi[$, we have
\begin{align*}
\lim_{(x,y)\to (0,0)}\left|1+xy^{2}\right|^\frac{1}{x^{2}+y^{2}}&=\lim_{r\to 0}|1+r^{3}\cos\theta\sin \theta|^{\frac{1}{r^{2}}},\\
&=\lim_{r\to 0}e^{\log |1+r^{3}\cos\theta\sin \theta|^{\frac{1}{r^{2}}}},\\
&=e^{\displaystyle \lim_{r\to 0}\frac{\log| 1+r^{3}\cos \theta\sin \theta|\underset{r\to 0}{\longrightarrow 0}}{r^{2}\underset{r\to 0}{\longrightarrow 0}}},\quad \text{(L'Hôpital's rule)}\\
&=e^{0},\\
&=1
\end{align*}
Therefore $\displaystyle \lim_{(x,y)\to (0,0)}f(x,y)=f(0,0)$ then $f$ is continuous function in $(0,0)$.
A: I think I managed to prove that the function is indeed continuous at $(0,0)$.
Choosing $y=\sqrt x$ , $x\rightarrow0$
I was able to prove the following:
$$
\lim_{(x,y)\rightarrow(0,0)}|1+xy^2|^\dfrac{1}{x^2+y^2}=\lim_{x\rightarrow0}(1+x^2)^\dfrac{1}{x^2+x}=\lim_{x\rightarrow0}((1+x^2)^\dfrac{1}{x^2})^\dfrac{x^2}{x^2+x}=e^0=1=f(0,0)
$$
which proves $f$ is continuous in $(0,0)$
I'm not quite sure this is enough, though all similar exercises I've seen in this subject are pretty much the same like this.
