Limit $\lim_{n \to \infty} \left( 1 - e^{- t} \right)^n$

The limit ($t \in \mathbb{R}$):

$$ \lim_{n \to \infty} \left( 1 - e^{- t} \right)^n $$

according to Wolfram Alpha is zero if $log(1 - cosh(t) + sinh(t)) < 0$. I didn't get why; if I see the step by step solution the result is infinity. Which is a way to get the right result (not in the software, but on a paper)?

$x = 1-e^{-t}$ is strictly less than one, so that $$ \lim_{n \to \infty} (1-e^{-t})^n = \lim_{n \to \infty} x^n $$ exists if and only if $$ x > -1 \iff t > -\log(2) $$ and in that case the limit is zero.

The solution from WA can be interpreted as follows: If $$ \log(1 - \cosh(t) + \sinh(t)) = \log(1-e^{-t}) $$ is negative then $$ \exp(\infty \cdot \log(1-e^{-t})) = \exp(-\infty) = \lim_{x \to -\infty}e^x = 0 \, . $$ But that is of limited use because

• it does not cover the case $t\le 0$, where $\log(1-e^{-t})$ is not defined, and
• $\log(1-e^{-t}) < \log(1) = 0$ holds for all real $t > 0$.