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

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Answer 1 · 2016-03-04T10:49:29

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

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

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

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

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

$= -oo$

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

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