Find the asymptotes of $$ \lim_{x \to \infty}x\cdot\exp\left(\dfrac{2}{x}\right)+1. $$ How is it done?
Finding asymptotes of exponential function and one-sided limit
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$\begingroup$
limits
exponential-function
3 Answers
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A related problem. We will use the Taylor series of the function $e^t$ at the point $t=0$,
$$ e^t = 1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\dots .$$
$$ x\,e^{2/x}+1 = x ( 1+\frac{2}{x}+ \frac{1}{2!}\frac{2^2}{x^2}+\dots )+1=x+3+\frac{2^2}{2!}\frac{1}{x}+\frac{2^3}{3!}\frac{1}{x^2}+\dots$$
$$ = x+3+O(1/x).$$
Now, you can see when $x$ goes to infinity, then you have
$$ x\,e^{2/x}+1 \sim x+3 $$
Here is the plot of $x\,e^{2/x}+1$ and the Oblique asymptote $x+3$
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0Could you further explain the second step? – 2012-11-28
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0@Grant: Note that he cited that $x+3$ is an Oblique asymptote. – 2012-11-28
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0@Grant: See the edit. – 2012-11-28
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1@BabakSorouh: Thanks for answering him. – 2012-11-28
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0@Grant: Try to plot the two functions $x\,e^{2/x}+1,\, x+3$ on the same graph and see what happens when x gets bigger. – 2012-11-28
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0Thank you for your effort, but I still don't understand how can you transform a simple expression ( x*e^(2/x) ) to some kind of series. – 2012-11-28
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0Aha, Taylor series! Thank you. – 2012-11-28
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0@Mhenni Why are the first two terms of the taylor series expansion enough to find the asymptote? Certainly there are functions where the first two terms wouldn't suffice to give an accurate approximation. – 2013-05-14
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There is an vertical asymptote for the function when $x\to0^+$.
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$$\lim_{x \to \infty}\frac d{dx}\left( x\cdot\exp\left(\dfrac{2}{x}\right)+1\right)=\lim_{x \to \infty}\exp\left(\frac2x\right)-\frac{2\exp\left(\frac2x\right)}{x}=1$$ therefore your function rises like $x$ asymptotically.