Given the set $D:=\{(x,y) \in \mathbb{R}^2: x > -1\}$ and the function $f: D\rightarrow \mathbb{R}$ through $f(x,y)= \begin{cases} \frac{y\ln(x+1)}{y^2+(\ln(x+1))^2} &\text{if $y \neq 0$ }\\0&\text{if $ y=0$}\end{cases}$
(i) Show that $\lim_{y\to 0} f (1,y) =0$ and $\lim_{x\to 1} f (x,0) =0$
(ii) Show that $\lim_{n\to \infty}(\frac{1}{n}, \ln (\frac{1}{n}+1))=(0,0)$
(iii) Show that $f$ is not continuous in $(0,0)$
To (i)
$\lim_{x\to 1} f (x,0) =0$ follows by definition, since $y=0$.
$\lim_{y\to 0} f (1,y) = \frac{0}{(\ln(2))^2}=0$
To (ii)
$\lim_{n\to \infty} \frac{1}{n}=0$
$\lim_{n\to \infty} \ln (\frac{1}{n}+1) = 0$
To (iii)
It is $X_m\to X_0\iff F(X_m)\to F(X_0)$
$X_m:=(\frac{1}{n}, \ln (\frac{1}{n}+1))$ converges $(o,o)$ But $f((X_m) = \frac{1}{2} \neq F(0,0)=0 \Rightarrow f$ is not continuous in $(0,0)$