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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?

3 Answers 3

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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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    @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.