Evaluate the following limit: `underset( x rarr 2) ( "lim")(x^4-16)/(x-2)`

Explanation

Given,

`(lim)_(x->2)(x^4−16)/(x−2)`

`(lim)_(x->2)(2^4−16)/(2−2)`

which is `0/0` form So,

by l-hospital rule

`(lim)_(x->2)(x^4−16)/(x−2)`

`=(lim)_(x->2)(4x^3)/(x)`

`=(lim)_(x->2)(4x^2)`

...