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I am finding (and computing the limits) of all vertical and horizontal asymptotes of ${5(e^x) \over (e^x)-2}$. I have found the vertical asymptote at $x=ln2$. I have also found the horizontal asymptote at y=5 by calculating the limit of the function at positive infinity. I know that to find the other horizontal asymptote (which is at $y=0$, based on the graph), I need to calculate the limit at negative infinity, but I cannot get it to equal zero.

my calculations thus far: lim as x approaches negative infinity of the function is equal to the limit as x approaches infinity of ${5 \over (e^x)-2}$, which I found by dividing both numerator and denominator by $e^x$.

but I just keep getting that the limit at negative infinity is the same as the limit at positive infinity - ${5\over(1-0)}$ or 5 - which I know is not correct.

what am i missing?

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Let $y=-x$; then you can more easily see that

$$\lim_{x\to-\infty}e^x=\lim_{y\to+\infty}e^{-y}=\lim_{y\to+\infty}\frac1{e^y}=0\;,$$

since the denominator $e^y$ is blowing up rather explosively as $y\to\infty$.

Going back to the original problem,

$$\lim_{x\to-\infty}\frac{5e^x}{e^x-2}=\frac{\lim\limits_{x\to-\infty}5e^x}{\lim\limits_{x\to-\infty}e^x-2}=\frac0{0-2}=0\;.$$

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  • $\begingroup$ okay, I understand the concept - that the limit of 5e^x or e^x is 0 as either approaches negative infinity - but why could I not get that with the algebra I did (namely dividing numerator and denominator by e^x, which is how I thought you solved infinity limits)? $\endgroup$ – theialux Sep 30 '12 at 1:54
  • $\begingroup$ @theialux: It isn’t always necessary, and this is a case in which it isn’t. You can do it that way, but then you have to evaluate correctly the resulting limit: in $\lim\limits_{x\to-\infty}\frac5{1-\frac2{e^x}}$ the denominator $\to-\infty$, so the whole thing $\to 0$. $\endgroup$ – Brian M. Scott Sep 30 '12 at 2:11
  • $\begingroup$ alright, thanks. $\endgroup$ – theialux Sep 30 '12 at 2:19
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Let us begin with the calculation: $${5(e^x) \over e^x-2}={5(e^x-2)+`10 \over e^x-2}=5+{`10 \over e^x-2}$$ Therefore, as $$x\rightarrow -\infty$$, $$e^x\rightarrow 0$$, and $$5+{`10 \over e^x-2}\rightarrow 5+ {10\over0-2}=5-5=0.$$

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