1. **Problem:** Check the continuity of the function $$f_1(x,y) = \begin{cases} \frac{x^2 + y^2}{\sqrt{x^2 + y^2 + 1} - 1} & (x,y) \neq (0,0) \\ 0 & (x,y) = (0,0) \end{cases}$$ at the point $(0,0)$. 2. **Recall:** A function $f(x,y)$ is continuous at $(0,0)$ if $$\lim_{(x,y) \to (0,0)} f(x,y) = f(0,0).$$ 3. **Evaluate the limit:** Let $r = \sqrt{x^2 + y^2}$. Then $$f_1(x,y) = \frac{r^2}{\sqrt{r^2 + 1} - 1}.$$ 4. **Simplify the expression:** Multiply numerator and denominator by the conjugate: $$\frac{r^2}{\sqrt{r^2 + 1} - 1} \cdot \frac{\sqrt{r^2 + 1} + 1}{\sqrt{r^2 + 1} + 1} = \frac{r^2 (\sqrt{r^2 + 1} + 1)}{(\sqrt{r^2 + 1})^2 - 1^2} = \frac{r^2 (\sqrt{r^2 + 1} + 1)}{r^2} = \sqrt{r^2 + 1} + 1.$$ 5. **Take the limit as $r \to 0$:** $$\lim_{r \to 0} f_1(x,y) = \lim_{r \to 0} (\sqrt{r^2 + 1} + 1) = \sqrt{0 + 1} + 1 = 1 + 1 = 2.$$ 6. **Compare with $f_1(0,0)$:** Given $f_1(0,0) = 0$, but the limit is 2, so $$\lim_{(x,y) \to (0,0)} f_1(x,y) \neq f_1(0,0).$$ 7. **Conclusion:** $f_1$ is **not continuous** at $(0,0)$. --- 1. **Problem:** Check the continuity of the function $$f_2(x,y) = \begin{cases} \frac{(1 + x + y) \sin y}{y} & (x,y) \neq (0,0) \\ 0 & (x,y) = (0,0) \end{cases}$$ at the point $(0,0)$. 2. **Recall:** Continuity requires $$\lim_{(x,y) \to (0,0)} f_2(x,y) = f_2(0,0) = 0.$$ 3. **Evaluate the limit:** For $y \neq 0$, $$f_2(x,y) = (1 + x + y) \frac{\sin y}{y}.$$ 4. **Use limit properties:** As $(x,y) \to (0,0)$, - $1 + x + y \to 1$ - $\frac{\sin y}{y} \to 1$ 5. **Therefore,** $$\lim_{(x,y) \to (0,0)} f_2(x,y) = 1 \times 1 = 1.$$ 6. **Compare with $f_2(0,0)$:** Given $f_2(0,0) = 0$, but the limit is 1, so $$\lim_{(x,y) \to (0,0)} f_2(x,y) \neq f_2(0,0).$$ 7. **Conclusion:** $f_2$ is **not continuous** at $(0,0)$. **Final answers:** - $f_1$ is not continuous at $(0,0)$. - $f_2$ is not continuous at $(0,0)$.