Answer
$\lim\limits_{x \to \infty} f(x) = 5$
Work Step by Step
We can find $\lim\limits_{x \to \infty}\frac{10~e^x-27}{2~e^x}$:
$\lim\limits_{x \to \infty}\frac{10~e^x-27}{2~e^x} = \lim\limits_{x \to \infty}\frac{10~e^x}{2~e^x}-\lim\limits_{x \to \infty}\frac{27}{2~e^x} = 5-0=5$
Note that this function is always less than 5 as $x$ approaches $\infty$, but it gets closer and closer to 5.
We can find $\lim\limits_{x \to \infty}\frac{5\sqrt{x}}{\sqrt{x-1}}$:
$\lim\limits_{x \to \infty}\frac{5\sqrt{x}}{\sqrt{x-1}} = 5 \times \lim\limits_{x \to \infty}\frac{\sqrt{x}}{\sqrt{x-1}} = (5)(1) = 5$
Note that this function is always greater than 5 as $x$ approaches $\infty$, but it gets closer and closer to 5.
Since $\frac{10~e^x-27}{2~e^x} \lt f(x) \lt \frac{5\sqrt{x}}{\sqrt{x-1}}$ for all $x \gt 1$, then $\lim\limits_{x \to \infty} f(x) = 5$
