Consider the function $B(t) = \frac {3e^t - 5}{e^t + 1}$.

As $t \to \infty$, the exponential pieces will dominate the constant pieces.

$$B(t) = \frac {3e^t - 5}{e^t + 1} \sim \frac {3 e^t}{e^t} = 3$$

As $t \to -\infty$, we have a different story. $\lim \limits _{t \to -\infty } e^t = 0$. Therefore, in this case, the constant terms dominate.

$$B(t) = \frac {3e^t - 5}{e^t + 1} \sim \frac {-5}{1} = -5$$

The graph has two different horizontal asymptotes.

Both the functions $S(t) = \sqrt {2t^3 - 5}$ and $L(t) = t\sqrt {9t+4}$ approach $\infty$ as $t \to \infty$.

$R(t) = \frac {S(t)}{L(t)} = \frac {\sqrt {2t^3 - 5}}{t\sqrt {9t+4}}$ is an indeterminate form. It is approaching the form $\frac {\infty }{\infty }$

If we examine the quotient in terms of dominance, then we think like...

$$R(t) = \frac {\sqrt {2t^3 - 5}}{t \sqrt {9t+4}} \sim \frac {\sqrt {2t^3}}{t \sqrt {9t}} = \frac {t \sqrt {2t}}{t \sqrt {9t}} = \frac {\sqrt {2}}{\sqrt {9}}$$

$$\lim \limits _{t \to \infty } R(t) = \frac {\sqrt {2}}{3}$$

Both the functions $S(t) = \sin (t)$ and $L(t) = t$ approach $0$ as $t \to 0$.

Do they approach $0$ at the same rate?

Let’s investigate $R(t) = \frac {\sin (t)}{t}$

$\sin (t)$ and $t$ approach $0$ at exactly the same rate. Their quotient approaches $1$.

We have seen an even weirder indeterminate form.

Everyone knows that $1$ to any power equals $1$. Everyone knows that numbers greater than $1$ to any power get bigger.

What about something like $\left ( 1 + \frac {1}{x} \right )^x$ ?

The base is getting closer to $1$, but they are always greater than $1$. The power is growing larger.

We are heading to something like $1^{\infty }$.

And we have seen this function actually approaches the value $e \approx 2.71828$.

How far out in the domain do you need to go for the graph to be inside the interval $[2.7, 2.8]$?