Question

Check the continuity of the function at the indicated point
f(x,y) = (7 x2 y)/(x2+y2) if (x,y) not equal to (0,0) ,
f(x,y) = 0 if (x,y) equal to (0,0) .

Answer 1

Answer on Question #42377 – Math – Integral Calculus:

Check the continuity of the function:

$$f(x, y) = \begin{cases} \dfrac{7x^2y}{x^2 + y^2}, & (x, y) \neq (0, 0) \\ 0, & (x, y) = (0, 0) \end{cases}.$$

Solution.

We need to check the continuity of $f$ at every point $(x_0, y_0)$.

1) $(x_0, y_0) \neq (0, 0)$:

$$\lim_{\substack{x\to x_{0}\\ y\to y_{0}}}f(x,y) = \lim_{\substack{x\to x_{0}\\ y\to y_{0}}}\dfrac{7x^{2}y}{x^{2} + y^{2}} = \dfrac{7x_{0}^{2}y_{0}}{x_{0}^{2} + y_{0}^{2}} = f(x_{0},y_{0});$$

So, $f$ is continuous at every point $(x_0, y_0) \neq (0, 0)$.

2) $(x_0, y_0) = (0, 0)$:

$$\lim_{\substack{x\to 0\\ y\to 0}}f(x,y) = \lim_{\substack{x\to 0\\ y\to 0}}\dfrac{7x^{2}y}{x^{2} + y^{2}} = \lim_{\substack{x\to 0\\ y\to 0}}\dfrac{7y}{1 + \left(\dfrac{y}{x}\right)^{2}};$$ $$\lim_{\substack{x\to 0\\ y\to 0}}\left|\dfrac{7y}{1 + \left(\dfrac{y}{x}\right)^{2}}\right| \leq \lim_{\substack{x\to 0\\ y\to 0}}\left|\dfrac{7y}{1}\right| = 7\lim_{\substack{x\to 0\\ y\to 0}}|y| = 0 \Rightarrow \lim_{\substack{x\to 0\\ y\to 0}}\dfrac{7y}{1 + \left(\dfrac{y}{x}\right)^{2}} = 0;$$

Hence:

$$\lim_{\substack{x\to 0\\ y\to 0}}f(x,y) = 0 = f(0,0);$$

So, $f$ is continuous at $(0,0)$.

We conclude that $f$ is continuous in $\mathbb{R} \times \mathbb{R}$.

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