` lim_(h to 0) (log (1+2h) - 2log (1+h))/h^(2) =...`

Answer

`-1`

Explanation

`underset (h to 0) lim ( loh (1+2h)-2log(1+h))/h^(2) " "[0/0" form"]`
Applying L'Hospital's rule, we get
`underset( h to 0) lim(2/(1+2h)-2/(1+h))/(2h) `
`underset( h to 0 ) lim (2+2h-2-4h)/(2h(1+2h)(1+h))`
` underset( h to 0) lim (-1)/((1+2h)(1+h)) = - 1`