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Find all asymptotes of: $$f(x) = \frac{a + be^x}{ae^x+b}$$

The way I've been taught is that the $+a$ and $+b$ in the numerator and denominator respectively do not contribute when x tends to infinity, therefore are negligible. Left with $f(x) = \frac{be^x}{ae^x} = \frac{b}{a}$, $y = \frac{b}{a}$ is the only asymptote I was able to identify (through this method).

However, plotting the function with $a = 3$ and $b = 2$, there is clearly another horizontal asymptote, where $y = \frac{a}{b}$:

enter image description here

Is there any way I could've known about the second asymptote without graphing?

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    $\begingroup$ Yes, look at what happens when $x \to -\infty$. $\endgroup$ – fixedp Jul 28 '16 at 10:07
  • $\begingroup$ Don't the $e^{x}$ in the numerator and denominator cancel out, cancelling the $e^{-\infty}$ out? $\endgroup$ – StopReadingThisUsername Jul 28 '16 at 10:09
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    $\begingroup$ No. $e^{-x}$ (and consequently $ae^{-x}$ and $be^{-x}$) become very small, so they are negligible. So in this case, the limit as $x \to -\infty$ is $\frac{a}{b}$. $\endgroup$ – fixedp Jul 28 '16 at 10:13
  • $\begingroup$ I'm sorry, I do not understand; the $e^{-x}$ cancels out, leaving $y = f(x) = \frac{b}{a}$, doesn't it? $\endgroup$ – StopReadingThisUsername Jul 28 '16 at 10:19
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    $\begingroup$ How does the $e^{-x}$ cancel out exactly? $$\displaystyle{\lim_{x\to -\infty}}ae^{-x}=0=\displaystyle{\lim_{x\to -\infty}}be^{-x}$$ so as you take the limit as $x\to {-\infty}$ you may as well consider these terms zero. $\endgroup$ – fixedp Jul 28 '16 at 10:29
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Note that $\displaystyle f(x)=\frac{1}{f(-x)}$. So if $\displaystyle\lim_{x\to+\infty} f(x)=L\not=0$ then $\displaystyle\lim_{x\to-\infty} f(x)=\frac{1}{L}$.

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    $\begingroup$ Good catch ! (+1) $\endgroup$ – Claude Leibovici Jul 28 '16 at 10:35
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The line $y = L$ is a horizontal asymptote of a function $f$ if $$\lim_{x \to \infty} f(x) = L$$ or $$\lim_{x \to -\infty} f(x) = L$$ While rational functions have at most one horizontal asymptote, a function may have two different horizontal asymptotes.

Note that $$\lim_{x \to -\infty} e^x = 0$$ Hence, $$\lim_{x \to \infty} \frac{a + be^x}{ae^x + b} = \lim_{x \to \infty} \frac{a + be^x}{ae^x + b} \cdot \frac{e^{-x}}{e^{-x}} = \lim_{x \to \infty} = \frac{ae^{-x} + b}{a + be^{-x}} = \lim_{x \to -\infty} \frac{ae^x + b}{a + be^x} = \frac{b}{a}$$ provided that $a \neq 0$, and $$\lim_{x \to -\infty} \frac{a + be^x}{ae^x + b} = \frac{a}{b}$$ provided that $b \neq 0$.

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