Question

does ( f(x)=\frac{1 - e^{-x}}{x} ) have any vertical asymptotes?
yes, ( x = 0 )
yes, ( x = 1 )
yes, ( x=-1 )
no vertical asymptotes
yes, ( x = 0 ) and ( x = 1 )
6 multiple choice 3 points
what are the horizontal asymptotes of ( f(x)=\frac{1 - e^{-x}}{x} )?
( y = 0 ) only
( y = 0 ) and ( y = 1 )
( y=-1 ) and ( y = 1 )
( y = 1 ) only
no horizontal asymptotes

Explanation:

Step1: Check vertical asymptotes

A vertical asymptote occurs at \(x = a\) if \(\lim_{x
ightarrow a^{-}}f(x)=\pm\infty\) or \(\lim_{x
ightarrow a^{+}}f(x)=\pm\infty\).
For \(f(x)=\frac{1 - e^{-x}}{x}\), use L - H rule for \(\lim_{x
ightarrow0}\frac{1 - e^{-x}}{x}\).
Differentiate numerator and denominator: \(\lim_{x
ightarrow0}\frac{e^{-x}}{1}=1\). So \(x = 0\) is not a vertical asymptote.

Step2: Check horizontal asymptotes

For horizontal asymptotes, find \(\lim_{x
ightarrow\pm\infty}\frac{1 - e^{-x}}{x}\).
As \(x
ightarrow\infty\), \(\lim_{x
ightarrow\infty}\frac{1 - e^{-x}}{x}=\lim_{x
ightarrow\infty}\frac{1}{x}-\lim_{x
ightarrow\infty}\frac{e^{-x}}{x}=0 - 0 = 0\).
As \(x
ightarrow-\infty\), \(e^{-x}
ightarrow\infty\), \(\lim_{x
ightarrow-\infty}\frac{1 - e^{-x}}{x}=\lim_{x
ightarrow-\infty}\frac{1}{x}-\lim_{x
ightarrow-\infty}\frac{e^{-x}}{x}\).
\(\lim_{x
ightarrow-\infty}\frac{1}{x}=0\), and \(\lim_{x
ightarrow-\infty}\frac{e^{-x}}{x}\) (using L - H rule: \(\lim_{x
ightarrow-\infty}\frac{e^{-x}}{x}=\lim_{x
ightarrow-\infty}\frac{-e^{-x}}{1}=-\infty\)). But \(\frac{1 - e^{-x}}{x}=\frac{-(e^{-x}-1)}{x}\), and \(\lim_{x
ightarrow-\infty}\frac{1 - e^{-x}}{x}=0\) (since degree of numerator and denominator: numerator \(e^{-x}-1\) behaves like \(e^{-x}\) (exponential) and denominator \(x\) (linear), and for large negative \(x\), \(e^{-x}\) grows, but \(\frac{e^{-x}}{x}
ightarrow0\) as \(x
ightarrow-\infty\) (using L - H rule \(\lim_{x
ightarrow-\infty}\frac{e^{-x}}{x}=\lim_{x
ightarrow-\infty}\frac{-e^{-x}}{1}=0\)).

Answer:

No vertical asymptotes; \(y = 0\) only.