Chapter 10 - Section 10.3 - Limits and Continuity: Algebraic Viewpoint - Exercises - Page 719: 71

$+\infty$

When $t\rightarrow\infty,$ $3t=$ "Big", and $2^{3t}$="Big", so the numerator $\rightarrow +\infty$ In the denominator, $e^{-t}=$ has the form $k^{-\infty}=0\implies k^{Big\ negative }= Small$ So, $1+5.3e^{-t}\rightarrow 1+$ small $\rightarrow 1$ The form of the limit is $\displaystyle \pm\frac{\infty}{k}=\pm\infty = \frac{Big}{k}=Big$ Thus, the limit is $+\infty$