Originally Posted by

**o_O** I'm not quite understanding your question but ...

$\displaystyle \lim_{t \to 1^{-}} \left[5(x-1)^{\frac{4}{5}}\right]_{0}^{t}

\: = \: \lim_{t \to 1^{-}} \left(5(t - 1)^{\frac{4}{5}} - 5(0-1)^{\frac{4}{5}}\right) = 5(0-1)^{\frac{4}{5}} - 5(-1)^{\frac{4}{5}} = 5 - 5 = 0$

$\displaystyle \lim_{t \to 0^{-}} \bigg[ \ln \left| e^{x} - 1 \right| \bigg]_{0}^{t} = \lim_{t \to 0^{-}} \bigg( \ln \left| e^{t} - 1\right| - \ln \left| e^{0} - 1\right| \bigg) = \left(\ln 0 - \ln 0\right) = -\infty - (- \infty) $

The latter is an indeterminate form but it does converge.

As for the $\displaystyle 1^{-}$, you don't really need it in this case but there are cases in which the limit as you approach a number from the left side is different from approaching it the right side.