Answer
$\dfrac{2}{5}$
Work Step by Step
Our aim is to compute the value of $\lim\limits_{x \to +\infty} (\dfrac{2}{5+4e^{-3x}})$.
Here, we need to use the formula of:
$\lim \limits_{x\to a}f(x)=f(a)$, this implies that:
$\lim\limits_{x \to +\infty} (\dfrac{2}{5+4e^{-3x}})=\dfrac{2}{5+4e^{-3x}} \\=\dfrac{2}{5+4e^{-3(+\infty)}}\\=\dfrac{2}{5+0} \\=\dfrac{2}{5}$
