As x->0-, the denominator 2(1-cos(x)) approaches 2(1-cos(0)) = 0.

Since we are given that lim x->0- of ((x-h)²*(x+1))/(2(1-cos(x)) exists, the only way this could possibly happen is if the numerator (x-h)²*(x+1) also approaches 0 as lim x->0-.

Note that as lim x->0-, (x-h)²*(x+1) goes to (-h)²*(1) = h², so h² = 0 which gives h = 0.

Now we would like to show that, when h = 0,

lim x->0- of ((x-h)²*(x+1))/(2(1-cos(x)))=1.

lim x->0- of (x²*(x+1))/(2(1-cos(x))

= lim x->0- of x²*(x+1)(1+cos(x)) / [2(1-cos(x))(1+cos(x))]

= lim x->0- of x²*(x+1)(1+cos(x)) / [2(1-cos^2(x))]

= lim x->0- of x²*(x+1)(1+cos(x)) / (2sin^2(x))

= (1/2) [lim x->0- of (x+1)(1+cos(x))] * [lim x->0- of x² / sin^2(x)]

= (1/2) [lim x->0- of (x+1)(1+cos(x))] / [lim x->0- of (sin(x) / x)]^2

= (1/2)(0+1)(1+cos(0)) / (1^2), since sin(x) / x goes to 1 as x goes to 0

= (1/2)(1)(2)

= 1.

So h = 0 actually is a solution, and there are no other solutions for h.

The answer is precisely h = 0.

Lord bless you today!