Question

How do you evaluate the limit of `lim (-4x^2-3x+6)` as `x->2`?

Answer 1

`lim_(x rarr 2) (-4x^2-3x+6)` = `-16`

Explanation:

`lim_(x rarr 2) (-4x^2-3x+6)`

`= (-4*2^2 - 3*2 + 6)`

`=(-16 -6 +6)`

`=-16`
The function is continuous at x = 2, where the function evaluates to
`-16`
and `lim_(x rarr 2^-)(-4x^2-3x+6)` = `lim_(x rarr 2 ^+)(-4x^2-3x+6)` = `-16`

Hence Proved.

graph{-4x^2 -3x + 6 [1.092, 2.778, -16.412, -15.569]}