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Homework Statement
Find \frac{\partial}{\partial x} if:
f(x,y) = \begin{cases}x^2\frac{\sin y}{y}, & y\neq 0\\0, &y=0 \end{cases}
Homework Equations
The Attempt at a Solution
If y\neq 0, then it's simple, but I get confused about the second part. How can I exactly utilize the limit definition of a derivative when I'm told that y=0?
f_x = \frac{\sin y}{y}\lim_{t\to 0}\frac{(x+t)^2 - x^2}{t} = 2x\frac{\sin y}{y}. If y = 0 can I say that \frac{\sin y}{y} = 1 since that would be the limit as y approaches 0, but y is already fixed at 0. Does the partial derivative not exist if y = 0?
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