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Math Help - Limit with indeterminate form

  1. #1
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    Limit with indeterminate form

    \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 {t^2+t}, but I don't know really where to go from there.
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  2. #2
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    Re: Limit with indeterminate form

    Quote Originally Posted by AZach View Post
    \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 {t^2+t}, but I don't know really where to go from there.
    \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*}
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  3. #3
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    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.
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