Investigate limℎ→0 1−cosℎ/ℎ2 numerically (and graphically if you have a graphing utility). Then prove that the limit is equal to 12.

Answer 1

From the graph, the limit equals 1/2

Proof:

lim_h→0(1-cos h)/h² = lim_h→0((1-cos h)/h²)(1+cos h)/(1+cos h)

= lim_h→0(1-cos² h)/h² (1/(1+cos h)) = lim_h→0(sin(h)/h)²lim_h→0(1/(1+cos h))

=(1)²(1/(1+1)) = 1/2