Evaluate the following limit: `underset( x rarr 0)((1+x)^6-1)/((1+x)^2-1)`

Explanation

Given limit: `lim_(x->0)((1+x)^6-1)/((1+x)^2-1)`
Substituting `0` as we get an indeterminant form of `(0/0)`
As it has come back as `(0/0)` form , so we have a tendency to use L’Hospitals Rule.
now by victimization L’Hospitals Rule the limit is given by,
`l=lim_(x->0)(6(1+x)^5-1)/(2(1-x))`
`=6/2`
`=3`