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Find the asymptotes of $$ \lim_{x \to \infty}x\cdot\exp\left(\dfrac{2}{x}\right)+1. $$ How is it done?

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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 $x+3$

enter image description here

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  • $\begingroup$ Could you further explain the second step? $\endgroup$ – user45045 Nov 28 '12 at 9:42
  • $\begingroup$ @Grant: Note that he cited that $x+3$ is an Oblique asymptote. $\endgroup$ – mrs Nov 28 '12 at 10:06
  • $\begingroup$ @Grant: See the edit. $\endgroup$ – Mhenni Benghorbal Nov 28 '12 at 10:14
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    $\begingroup$ @BabakSorouh: Thanks for answering him. $\endgroup$ – Mhenni Benghorbal Nov 28 '12 at 10:14
  • $\begingroup$ @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. $\endgroup$ – Mhenni Benghorbal Nov 28 '12 at 10:17
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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.

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