why does e raised to the power of negative infinity equal 0? Why is it that e raised to the power of negative infinity would equal 0 instead of negative infinity? I am working on problems with regards to limits of integration, specifically improper integrals and a little confused as to what things approach infinity our negative infinity versus approaching zero
Note that: $e^{k} = \dfrac{1}{e^{-k}}$. Thus if $k \to -\infty \Rightarrow -k \to +\infty$, and $e^{-k} \to +\infty$ because $e^{-k} &gt; -k$, and the as the latter approaches $+\infty$, so does the former, hence $e^k=\dfrac{1}{e^{-k}} \to 0$.
$$e^{-k} = \dfrac{1}{e^{k}}$$
Also $e^{\infty} = \infty$  and  $e^{-k}= \dfrac{1}{e^k} \Rightarrow e^{-\infty} = \dfrac{1}{\infty}= 0$

(anything divided by infinity is zero)