Question

Prove that the function\(f\] is the function of two variables that is differentiable at\((a,b),\]then\(f\] is continuous at\((a,b).\]

Short answer

The answer is stated below.

Step-by-step solution

Definition 7.

“If\(z = f(x,y){\rm{,}}\]then\(f\]is differentiable at\((a,b)\]if\(\Delta z\]can be expressed in the form\(\Delta z = {f_x}(a,b)\Delta x + {f_y}(a,b)\Delta y + {\varepsilon _1}\Delta x + {\varepsilon _2}\Delta y,\]where\({\varepsilon _1}\]and\({\varepsilon _2} \to 0.{\rm{ as }}(\Delta x,\Delta y) \to (0,0).\]”

Use the definition 7 for calculation.

The function is differentiable at\((a,b).\]

Let the function be\({\rm{z = f(x, y)}}.\]

The function\({\rm{f(x, y)}}\] is differentiable at\((a,b).\]

To show that the function\({\rm{f(x, y)}}\]is continuous at\((a,b),\]it is enough to prove that,

\(\mathop {\lim }\limits_{(\Delta x,\Delta y) \to (0,0)} f(a + \Delta x,b + \Delta y) = f(a,b)\]

By using definition 7,

\(\begin{aligned}{l}f(a + \Delta x,b + \Delta y) - f(a,b) = \Delta z = {f_x}(a,b)\Delta x + {f_y}(a,b)\Delta y + {\varepsilon _1}\Delta x + {\varepsilon _2}\Delta y\\f(a + \Delta x,b + \Delta y) = f(a,b) + {f_x}(a,b)\Delta x + {f_y}(a,b)\Delta y + {\varepsilon _1}\Delta x + {\varepsilon _2}\Delta y\end{aligned}\]

Taking limit on both sides as\((\Delta x,\Delta y) \to (0,0),\]

\(\begin{aligned}{l}f(a + \Delta x,b + \Delta y) = f(a,b) + {f_x}(a,b)\Delta x + {f_y}(a,b)\Delta y + {\varepsilon _1}\Delta x + {\varepsilon _2}\Delta y\\\mathop {\lim }\limits_{(\Delta x,\Delta y) \to (0,0)} f(a + \Delta x,b + \Delta y)\\ = \mathop {\lim }\limits_{(\Delta x,\Delta y) \to (0,0)} \left( {f(a,b) + {f_x}(a,b)\Delta x + {f_y}(a,b)\Delta y + {\varepsilon _1}\Delta x + {\varepsilon _2}\Delta y} \right)\\\mathop {\lim }\limits_{(\Delta x,\Delta y) \to (0,0)} f(a + \Delta x,b + \Delta y)\\ = \left\{ {\begin{aligned}{*{20}{l}}{\mathop {\lim }\limits_{(\Delta x,\Delta y) \to (0,0)} f(a,b) + \mathop {\lim }\limits_{(\Delta x,\Delta y) \to (0,0)} {f_x}(a,b)\Delta x + }\\{\mathop {\lim }\limits_{(\Delta x,\Delta y) \to (0,0)} {f_y}(a,b)\Delta y + \mathop {\lim }\limits_{(\Delta x,\Delta y) \to (0,0)} {\varepsilon _1}\Delta x + \mathop {\lim }\limits_{(\Delta x,\Delta y) \to (0,0)} {\varepsilon _2}\Delta y}\end{aligned}} \right\}\end{aligned}\]

If\((\Delta x,\Delta y) \to (0,0),\]then\({\varepsilon _1}\]and\({\varepsilon _2} \to 0.\]

\(\begin{aligned}{l}\mathop {\lim }\limits_{(\Delta x,\Delta y) \to (0,0)} f(a + \Delta x,b + \Delta y)\\ = \left\{ {\begin{aligned}{*{20}{c}}{\mathop {\lim }\limits_{(\Delta x,\Delta y) \to (0,0)} f(a,b) + \mathop {\lim }\limits_{(\Delta x,\Delta y) \to (0,0)} {f_x}(a,b)\Delta x + }\\{\mathop {\lim }\limits_{(\Delta x,\Delta y) \to (0,0)} {f_y}(a,b)\Delta y + \mathop {\lim }\limits_{(\Delta x,\Delta y) \to (0,0)} {\varepsilon _1}\Delta x + \mathop {\lim }\limits_{(\Delta x,\Delta y) \to (0,0)} {\varepsilon _2}\Delta y}\end{aligned}} \right\}\mathop {\lim }\limits_{(\Delta x,\Delta y) \to (0,0)} f(a + \Delta x,b + \Delta y) = f(a,b)\end{aligned}\]

Hence, the function\({\rm{f(x, y)}}\]is continuous at\((a,b).\]