Answer
$+\infty$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to + \infty} \dfrac{e^{3x}}{x^2}$.
But $\lim\limits_{x \to + \infty} \dfrac{e^{3x}}{x^2}=\dfrac{\infty}{\infty}$
We can see that the numerator and denominator have a limit of $\infty$, so the limit shows the indeterminate form of type $\dfrac{\infty}{\infty}$. So, we will apply L'Hopital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{A(x) }{B(x)}=\lim\limits_{x \to a}\dfrac{A'(x)}{B'(x)}$
where, $a$ can be any real number, infinity or negative infinity.
$\lim\limits_{x \to + \infty} \dfrac{e^{3x}}{x^2}=\lim\limits_{x \to + \infty} \dfrac{3e^{3x}}{2x}=\dfrac{\infty}{\infty}$
Again apply L'Hopital's rule :
$\lim\limits_{x \to + \infty} \dfrac{3e^{3x}}{2x}=\lim\limits_{x \to + \infty} \dfrac{9e^{3x}}{2}=+\infty$