 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2012/f/f2/b/ii/solution

h^{'} (0) = - \frac{f _{x} ( 0 , 0 )}{f _{y} ( 0 , 0 )} = - 1.

All sufficiently near points of the curve have

y = h (x)

, and the only such point with

x = 0

is the origin itself. Therefore for the punctured approach along the curve,

(x, y) \to (0, 0) (x, y) \in C, (x, y) \neq = (0, 0) lim \frac{y}{x} = x \to 0 lim \frac{h ( x ) - h ( 0 )}{x} = h^{'} (0) = - 1.

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