Answer
$\displaystyle \lim_{t\rightarrow\infty}I(t)=\infty,\quad\lim_{t\rightarrow\infty}\frac{I(t)}{E(t)}=2.5$
Although $I(t)$ rises indefinitely, it remains $2.5$ times greater than $E(t)$.
Work Step by Step
$t$ is "approaching $+\infty$" means that $t$ is assuming positive values of greater and greater magnitude.
Observe the table below:
As $t$ assumes positive values of greater and greater magnitude,
the function value $I(t)$ rises without bound.
So, we write
$\displaystyle \lim_{t\rightarrow\infty}I(t)=\infty.$
The ratio, however, approaches the value $2.5$.
$\displaystyle \lim_{t\rightarrow\infty}\frac{I(t)}{E(t)}=2.5$
We see that although $I(t)$ rises indefinitely, it remains $2.5$ times greater than $E(t)$.
