Question

57 - 62. transcendental functions determine the end behavior of the following transcendental functions by analyzing appropriate limits. then provide a simple sketch of the associated graph, showing asymptotes if they exist.

1. $f(x)=-3e^{-x}$ 58. $f(x)=2^{x}$ 59. $f(x)=1 - ln x$
2. $f(x)=|ln x|$ 61. $f(x)=sin x$ 62. $f(x)=\frac{50}{e^{2x}}$

Explanation:

Step1: Analyze $\lim_{x

ightarrow+\infty}-3e^{-x}$
We know that the exponential - function $y = e^{-x}=\frac{1}{e^{x}}$. As $x
ightarrow+\infty$, $e^{x}
ightarrow+\infty$, so $\lim_{x
ightarrow+\infty}-3e^{-x}=\lim_{x
ightarrow+\infty}-\frac{3}{e^{x}} = 0$.

Step2: Analyze $\lim_{x

ightarrow-\infty}-3e^{-x}$
As $x
ightarrow-\infty$, $-x
ightarrow+\infty$, and $e^{-x}
ightarrow+\infty$. So $\lim_{x
ightarrow-\infty}-3e^{-x}=-\infty$.

Step3: Determine asymptotes

Since $\lim_{x
ightarrow+\infty}-3e^{-x}=0$, the horizontal asymptote is $y = 0$ as $x
ightarrow+\infty$. There are no vertical asymptotes for the function $y=-3e^{-x}$ because the domain of $y = - 3e^{-x}$ is $(-\infty,+\infty)$.

To sketch the graph:

• The function $y=-3e^{-x}$ is always negative because $e^{-x}>0$ for all real - valued $x$ and the coefficient $-3$ is negative.
• As $x$ increases, the function approaches the $x$ - axis ($y = 0$) from below. As $x$ decreases, the function decreases without bound.

Answer:

As $x
ightarrow+\infty$, $f(x)
ightarrow0$; as $x
ightarrow-\infty$, $f(x)
ightarrow-\infty$. The horizontal asymptote is $y = 0$ as $x
ightarrow+\infty$.