Question

Find the value of limit (cosah-cosbh)/h^2 as h approaches to 0?

Answer 1

Please see below.

Explanation:

`(cosah - cos bh)/h^2 * (cosah + cos bh)/ (cosah + cos bh) = (cos^2ah-cos^2bh)/(h^2(cosah + cos bh))`

Now use `cos^2 A = 1-sin^2A`, to get

`= (-sin^2 ah + sin^2bh)/(h^2(cosah + cos bh))`

`= ((-sin^2 ah)/h^2 + (sin^2bh)/h^2)1/(cosah + cos bh)`

`= [(-a^2((sin ah)/(ah))^2) + (b^2((sinbh)/(bh))^2)]1/(cosah + cos bh)`

The limit as `xrarr0` is

`[-a^2(1)+b^2(1)]1/(1+1) = (b^2-a^2)/2`