lim (x->0) 1-cos(x) / x^2?

Question

I can find this limit graphically ( .5 ) - and with a table -- but I can't figure out the analytical...
Can anyone point me in the right direction to prove that the limit = 1/2 ?
I appreciate it!

Answer 1

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

Answer 2

I assume you mean

Limit x→0 of [(1 - cosx) / x²]

This is the indeterminant form 0/0 so L'Hospital's rule applies. The rule says the limit of the quotient is equal to the limit of the derivative of the numerator divided by the deriviative of the denominator.

Limit x→0 of [(1 - cosx) / x²]

= Limit x→0 of [sinx / (2x)]

= Limit x→0 of [cosx / 2] = 1/2

Answer 3

lim x->0 cos(x)/x DNE lim (x-)->0 cos(x)/x = -(cos(0)/0) = -one million/0 = -? lim (x+)->0 cos(x)/x = (cos(0)/0) = one million/0 = ? The shrink returned itself would not exist. additionally, in case you advise lim x->0 (cos(x/x)) then it would honestly be cos(one million). the graph of cos(x/x) is a in the present day horizontal line that has a hollow at x=0.