y^3 - x^2 = 4
3y^2 y' - 2x = 0
y' = 2x/3y^2
6y y'^2 + 3y^2 y'' - 2 = 0
y'' = (2 - 6yy'^2)/3
= (2 - 6y(2x/3y^2)^2)/3
y = x^a + 1/x^a
y' = ax^(a-1) - a/x^(a+1)
y'' = a(a-1)x^(a-2) + a(a+1)/x^(a+2)
x^2 y'' = a(a-1)x^a + a(a+1)/x^a
xy' = ax^a - a/x^a
-a^2 y = -a^2 x^a - a^2/x^a
Add 'em up: 0
cos(x-y) = y(2x+1)^3
-sin(x-y)(1-y') = y'(2x+1)^3 + 6y(2x+1)^2
y'(sin(x-y) - y'(2x+1)^3 = xsin(x-y) + 6y(2x+1)^2
lim sin^2 x/x = 2sincos/1 = 0
f(0) = A + 3
f(0) = 5 + 0B = 5
So, if A=2, then the limit exists from both sides
thank u so much for ur effort and time
helped me alot!!!!!
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