Question 524957
The distance in feet that an object falls in a vacuum is given by 
s(t)= 16t^2, where t represents the time in seconds. 

a) Find
s(2+h) - s(2)/h
= {{{(16(2+h)^2 - 16*2^2)/h}}}
= {{{16*(4 + 4h + h^2 - 4)/h}}}
= {{{16*(4h + h^2)/h}}}
= {{{16*(4 + h)}}}
= 64 + 16h
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b) what happens when h tends to 0 ?
s(2+h) - s(2)/h approaches 64