# f(x,y) = 2x^3xy^25x^2y^2 # Let us find the first partial derivatives # (partial f) / (partial x) = 6x^2y^210x# # (partial f) / (partial y) = 2xy2y# So our critical equations are # 6x^2y^210x = 0# # 2xy2y = 0# From the second equation we have # 2y(x1) = 0 => x=1,y=0# Subs #x=1# into the First equation and we getFall 13 S Jamshidi 4 x4 y4 z4 =1 If x,y,z are nonzero, then we can consider Therefore, we have the following equations 1 1=2x2 2 1=2y2 3 1=2z2 4 x4 y4 z4 =1 Remember, we can only make this simplification if all the variables are nonzero!X2 sin2 y x2 2y2 We probe the limit by the straightline approaches y = mx which gives lim (x,y)→(0,0) x 2sin y x2 2y 2 = lim x→0 x 2sin2 mx x 2m2x2 = lim x→0 sin mx 12m2 = 0 We might suspect that the limit exists and is equal to 0 To justify this, we notice that since 0 ≤ x2 x2 2y2 ≤ 1, we have the inequalities 0 ≤ x 2sin y
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