Answer
Using the definition of differentiability in Section 15.4, we show that:
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{f\left( {x,y} \right)}}{{\sqrt {{x^2} + {y^2}} }} = 0$
Work Step by Step
We are given $f\left( {0,0} \right) = {f_x}\left( {0,0} \right) = {f_y}\left( {0,0} \right) = 0$.
If $f\left( {x,y} \right)$ is differentiable at $\left( {0,0} \right)$, then by definition of differentiability in Section 15.4, $f\left( {x,y} \right)$ is locally linear at $\left( {0,0} \right)$. That is,
$f\left( {x,y} \right) = L\left( {x,y} \right) + e\left( {x,y} \right)$,
where $L\left( {x,y} \right) = f\left( {0,0} \right) + {f_x}\left( {0,0} \right)x + {f_y}\left( {0,0} \right)y$ and $e\left( {x,y} \right)$ satisfies
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{e\left( {x,y} \right)}}{{\sqrt {{x^2} + {y^2}} }} = 0$
Next, we evaluate $\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{f\left( {x,y} \right)}}{{\sqrt {{x^2} + {y^2}} }}$
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{f\left( {x,y} \right)}}{{\sqrt {{x^2} + {y^2}} }} = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{L\left( {x,y} \right) + e\left( {x,y} \right)}}{{\sqrt {{x^2} + {y^2}} }}$
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{f\left( {x,y} \right)}}{{\sqrt {{x^2} + {y^2}} }} = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{L\left( {x,y} \right)}}{{\sqrt {{x^2} + {y^2}} }} + \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{e\left( {x,y} \right)}}{{\sqrt {{x^2} + {y^2}} }}$
1. Consider the first limit on the right-hand side:
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{L\left( {x,y} \right)}}{{\sqrt {{x^2} + {y^2}} }}$
Since $f\left( {0,0} \right) = {f_x}\left( {0,0} \right) = {f_y}\left( {0,0} \right) = 0{\rm{, so}}L\left( {x,y} \right) = 0$. Thus
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{L\left( {x,y} \right)}}{{\sqrt {{x^2} + {y^2}} }} = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{0}{{\sqrt {{x^2} + {y^2}} }} = 0$
2. Consider the second limit on the right-hand side:
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{e\left( {x,y} \right)}}{{\sqrt {{x^2} + {y^2}} }}$
Since $f\left( {x,y} \right)$ is locally linear at $\left( {0,0} \right)$,
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{e\left( {x,y} \right)}}{{\sqrt {{x^2} + {y^2}} }} = 0$
Therefore, $\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{f\left( {x,y} \right)}}{{\sqrt {{x^2} + {y^2}} }} = 0$.
