Question

How do you find the limit of `[(1/t)-(1/t^2+t)]` as t approaches 0?

Answer 1

`lim_{t -> 0}[1/t - (1/t^2 + t)] = -oo`

Explanation:

`lim_{t -> 0}[1/t - (1/t^2 + t)] = lim_{t -> 0}(1/t - 1/t^2 - t)`

`= lim_{t -> 0}((-t^3 + t - 1)/t^2)`

`= lim_{t -> 0}((-(0)^3 + (0) - 1)/t^2)`

`= lim_{t -> 0}((- 1)/t^2)`

`= -oo`

graph{1/x - 1/x^2 -x [-10, 10, -5, 5]}