Answer
Improper, converges to $\displaystyle \frac{1}{3}$.
Work Step by Step
The integral has form:
(Improper Integrals with Infinite Integration Limits)
2. If $f$ is continuous on the interval $(-\infty, b]$, then $\displaystyle \int_{-\infty}^{b}f(x)dx=\lim_{a\rightarrow-\infty}\int_{a}^{b}f(x)dx$.
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$\displaystyle \int_{-\infty}^{0}e^{3x}dx=\lim_{a\rightarrow-\infty}\int_{b}^{0}e^{3x}dx$
$=\displaystyle \lim_{a\rightarrow-\infty}[\frac{1}{3}e^{3x}]_{a}^{0}$
$=\displaystyle \lim_{a\rightarrow-\infty}[\frac{1}{3}-\frac{1}{3}e^{3a}]$
$=\displaystyle \frac{1}{3}-0$
Converges.
