Question

How do you evaluate the limit `2t^2+8t+8` as t approaches `2`?

Answer 1

`lim_(t rarr 2) (2t^2+8t+8) = 32`

Explanation:

If we define `f(t)=2t^2+8t+8`, then `f(t)` is continuous everywhere (ie it is well behaved and it has no jumps, discontinuities or places where the function is not defined).

Consequently

`lim_(t rarr a) f(t)= f(a)` for all values of `a`

hence

`lim_(t rarr 2) (2t^2+8t+8) = lim_(t rarr 2) f(t)`
`" " = f(2)`
`" " = 8+16+8`
`" " = 32`

Answer 2

`32`

Explanation:

The function is a polynomial and continuous thus the limit can be evaluated by substitution.

`lim_(t to2)(2t^2+8t+8)=8+16+8=32`