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Suppose f(x) = {a+bx, x<1,4,x=1,b-a,x, x>1} and if limx→1 f(x) = f(1), what are possible values of a and b?
Evaluate:
limx→1 [(x - 2)/(x2 - x) - 1/(x3 - 3x2 + 2x)]
If(x) = {mx2 + n, x < 0
mx + m, 0 ≤ x ≤ 1
nx2 + m, x > 1
For what integers m and n does both limx→0 f(x) and limx→1 f(x) exist?
Evaluate: limh→0 [((a + h)2 sin(a + h) - a2 sin a ) / h]
find limx → 1 f(x), where f(x) = {x2 - 1, x ≤ 1 -x2 - 1, x > 1
Evaluate: limx→2 [(1/x - 2) - (2(2x - 3)/x3 - 3x2 + 2x)]
Evaluate: limx→0 [(sin(2 + x) - sin(2 - x)) / x]
Evaluate: limx→0 [(tan x - sin x) / (sin3 x)]
Evaluate: limx→π/2 [(1 - sin x) / cos x]
Find the derivative of f(x) = (2x + 3)/(x - 2) from the first principle.
Evaluate:limx→1 (x15 - 1)/(x10 - 1)
Evaluate: limx→π/2 (secx - tanx)
Evaluate: limx→π/6 [(2sin2 x + sinx - 1) / (2sin2 x - 3sin x + 1)]
Evaluate: limx → 0 (sin 4x/sin 2x)