Question

Let f : \(\R^2 → \R\) be defined as follows :
\(f(x,y)=\begin{cases}    \frac{x^4y^3}{x^6+y^6} & \text{if }(x,y) \ne (0,0)\\     0  & \text{if } (x,y)=(0,0) \end{cases}\)
Then

• A. \(\lim\limits_{t \rightarrow 0}\frac{f(t,t)-f(0,0)}{t}\) exists and equals \(\frac{1}{2}\)
• B. \(\frac{∂f}{∂x}|_{(0,0)}\) exists and equals 0
• C. \(\frac{∂f}{∂y}|_{(0,0)}\) exists and equals 0
• D. \(\lim\limits_{t \rightarrow 0}\frac{f(t,2t)-f(0,0)}{t}\) exists and equals \(\frac{1}{3}\)

Answer 1

The Correct Option is A, B, C

To determine the behavior of the function \( f(x, y) \) at the origin and evaluate the limits and derivatives, we need to carefully analyze the given conditions. The function is defined as:

\(f(x,y)=\begin{cases} \frac{x^4y^3}{x^6+y^6} & \text{if }(x,y) \ne (0,0)\\ 0 & \text{if } (x,y)=(0,0) \end{cases}\)

1. We first verify \(\lim\limits_{t \rightarrow 0}\frac{f(t,t)-f(0,0)}{t}\):

   Plug in \( x = t \) and \( y = t \), then:

   \( f(t, t) = \frac{t^4 \cdot t^3}{t^6 + t^6} = \frac{t^7}{2t^6} = \frac{t}{2} \)

   Thus, \(\frac{f(t,t)-f(0,0)}{t} = \frac{\frac{t}{2} - 0}{t} = \frac{1}{2}\).

   Taking the limit as \( t \rightarrow 0 \), we find:

   \(\lim\limits_{t \rightarrow 0}\frac{f(t,t)-f(0,0)}{t} = \frac{1}{2}\).

2. Derivatives at \((0,0)\):
   • Partial derivative of \(f\) with respect to \(x\) at \((0,0)\):

     The partial derivative is defined as:

     \( \frac{∂f}{∂x}|_{(0,0)} = \lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} \)

     When \( y = 0 \), \( f(h,0) = 0 \), so:

     \( \frac{∂f}{∂x}|_{(0,0)} = \lim_{h \to 0} \frac{0 - 0}{h} = 0 \)

   • Partial derivative of \(f\) with respect to \(y\) at \((0,0)\):

     The partial derivative is defined as:

     \( \frac{∂f}{∂y}|_{(0,0)} = \lim_{k \to 0} \frac{f(0,k) - f(0,0)}{k} \)

     When \( x = 0 \), \( f(0,k) = 0 \), so:

     \( \frac{∂f}{∂y}|_{(0,0)} = \lim_{k \to 0} \frac{0 - 0}{k} = 0 \)

3. Finally, verify \(\lim\limits_{t \rightarrow 0}\frac{f(t,2t)-f(0,0)}{t}\):

   Set \(x = t\) and \(y = 2t\), then:

   \( f(t, 2t) = \frac{t^4 \cdot (2t)^3}{t^6 + (2t)^6} = \frac{8t^7}{t^6 + 64t^6} = \frac{8t}{65} \)

   Then, \(\frac{f(t,2t)-f(0,0)}{t} = \frac{\frac{8t}{65} - 0}{t} = \frac{8}{65}\).

From the above analysis, the correct options are:

• \(\lim\limits_{t \rightarrow 0}\frac{f(t,t)-f(0,0)}{t}\) exists and equals \(\frac{1}{2}\)
• \(\frac{∂f}{∂x}|_{(0,0)}\) exists and equals 0
• \(\frac{∂f}{∂y}|_{(0,0)}\) exists and equals 0