Limit with indeterminate form

$\displaystyle \lim \limits_{t \to 0}(\frac{1}{t}-\frac{1}{t^2+t})$

I wasn't sure what to do with this problem on a test and got it wrong. Substituting 0 directly into the problem didn't seem to help me as it became an indeterminate form. I know the answer is 1 now, but I still can't reverse evaluate the limit. I factored $\displaystyle {t^2+t}$, but I don't know really where to go from there.

Re: Limit with indeterminate form

Quote:

Originally Posted by

**AZach** $\displaystyle \lim \limits_{t \to 0}(\frac{1}{t}-\frac{1}{t^2+t})$

I wasn't sure what to do with this problem on a test and got it wrong. Substituting 0 directly into the problem didn't seem to help me as it became an indeterminate form. I know the answer is 1 now, but I still can't reverse evaluate the limit. I factored $\displaystyle {t^2+t}$, but I don't know really where to go from there.

$\displaystyle \displaystyle \begin{align*} \lim_{t \to 0} \left( \frac{1}{t} - \frac{1}{t^2 + t} \right) &= \lim_{t \to 0} \left[ \frac{t + 1}{t(t + 1)} - \frac{1}{t(t + 1)} \right] \\ &= \lim_{t \to 0} \left[ \frac{t}{t(t + 1)} \right] \\ &= \lim_{t \to 0}\left( \frac{1}{t + 1} \right) \\ &= \frac{1}{0 + 1} \\ &= 1 \end{align*}$

Re: Limit with indeterminate form

My laTex is bad today.

Make common denominator and factor out the t. You're left with 1/(t+1). Substitute.