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# How do you find the asymptotes for  f(x)=xe^-x?

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#### Explanation

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#### Explanation:

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Jim S Share
Jun 20, 2018

${C}_{f}$ has an horizontical asymptote $y = 0$ at $+ \infty$

#### Explanation:

$f \left(x\right) = x {e}^{- x} = \frac{x}{e} ^ x$

• $f$ has domain the set of real numbers $\mathbb{R}$ because ${e}^{x} > 0$ , $\forall x$$\in$$\mathbb{R}$

This means we're not looking for vertical asymptotes.

For oblique linear asymptotes (y=λx+b) we have:

${\lim}_{x \to - \infty} f \frac{x}{x} = {\lim}_{x \to - \infty} \frac{x {e}^{- x}}{x} = {\lim}_{x \to - \infty} {e}^{- x} =$

$+ \infty$

No oblique/horizontical asymptote at $- \infty$

lim_(xto+oo)f(x)/x=lim_(xto+oo)e^(-x)=^((e^(-oo)))0=λ

lim_(xto+oo)(f(x)-λx)=lim_(xto+oo)f(x)=lim_(xto+oo)xe^(-x)=

${\lim}_{x \to + \infty} \frac{x}{e} ^ x {=}_{D L H}^{\left(\frac{+ \infty}{+ \infty}\right)} {\lim}_{x \to + \infty} \frac{1}{e} ^ x {=}^{\left(\frac{1}{+ \infty}\right)} 0 = b$

As a result ${C}_{f}$ has an horizontical asymptote $y = 0$ at $+ \infty$

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