Question

lim h -> 0 ?

Answer 1

`9`

Explanation:

`f(x)=abs(-9x-7)=abs(9x+7)`

`(f(6+h)-f(6))/h=(abs(9(6+h)+7)-abs(9xx6+7))/h=`
`(abs(61+9h)-61)/h = 9((abs(61/9+h)-61/9)/h)`

now

a) if `h > 0->abs(61/9+h)=61/9+h` so
`9((abs(61/9+h)-61/9)/h)=9(61/9+h-61/9)/h=9`

b) if `h < 0->abs(61/9-h)=61/9-h` so
`9((abs(61/9+h)-61/9)/h)=9(61/9-h-61/9)/(-h)=9`

No limit is required.