# `lim_(t->0) (1/t - 1/(t^2 + t)) ` Evaluate the limit, if it exists.

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### 2 Answers

`lim_(t->0) (1/t - 1/(t^2 + t))`

sol:

`lim_(t->0) (1/t - 1/t(t + 1))`

=>`lim_(t->0) (1/t(1 - 1/(t + 1))`

=>`lim_(t->0) (1/t((t+1 - 1)/(t + 1))`

=>`lim_(t->0) (1/t((t)/(t + 1))`

=> `lim_(t->0) (1/(t + 1))`

on t->0

`lim_(t->0) (1/(t + 1)) = 1`

Combine the fraction so that it is in the form f(x)/g(x):

`lim_(t->0)( t^2 / ( t^3 + t^2 ) )` Now plugging in t results in the indeterminate form 0/0. Apply l'hopital's rule and take the derivative of both f(x) and g(x).

`lim_(t->0) ( (2t)/(3t^2 + 2t) )` Now plugging in t results in the indeterminate form 0/0. Again apply l'hopital's rule and take the derivative of the new f(x) and g(x).

`lim_(t->0) ( (2)/(6t + 2) )` Now plugging in t shows that the limit is 1.

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