Does $\lim_{x\to 0}\frac{x_1^2}{x_1+x_2}$ exist?

Question

Let $f(x):\mathbb{R}^2\rightarrow\mathbb{R}$ via $$f(x)=\begin{cases} \dfrac{x_1^2}{x_1+x_2}, &\text{ if }x_1+x_2\neq 0\\ 0, & \text{ otherwise } \end{cases}$$ Does $\lim_\limits{x\rightarrow 0}f(x)$ exist? I tried several sequences and it seems that the limit $=0$, but I haven't found a good way to prove it using $\varepsilon-\delta$.

Answer 1

Let $x_{1,n}=1/\sqrt{n}$, and $x_{2,n}=1/n^2-x_{1,n}$. Then $(x_{1,n},x_{2,n})\to 0$ as $n\to\infty$. But $f(x_{1,n},x_{2,n})=n\to\infty$.

Answer 2

Note that

• for $x_1=0\implies \frac{x_1^2}{x_1+x_2}\to 0$

• for $x_1=t \quad x_2=-t+t^2\implies \frac{x_1^2}{x_1+x_2}= \frac{t^2}{t-t+t^2} \to 1$

Answer 3

With $x_2=\lambda x_1^2-x_1$, it becomes

$$\lim_{x_1\to 0}\frac {x_1^2}{\lambda x_1^2}=\frac {1}{\lambda} $$

from here, the limit does not exist.