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Given the function, $f:\mathbb{R} \setminus {\{0}\}\to\mathbb{R}$, $f(x)=x+\frac{1}{x}$.

Find the limits of $f$ at infinity(and negative infinity) and all asymptotes of $f$.

I don't know where I have to start. Really, I have no idea.

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I don't why downvote? –  89085731 Mar 30 '12 at 10:55
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Please add your question to the post rather than only to the title. –  user21436 Mar 30 '12 at 11:45
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2 Answers

up vote 4 down vote accepted

We can see where is the horizontal asymptote by doing $\lim\limits_{x\to \infty}x+\frac{1}{x}$ and $\lim\limits_{x\to -\infty}x+\frac{1}{x}$

The vertical asymptote is where $\frac{1}{x}=0$ and to the limits with that number. In this case, $\infty \text{ and }-\infty$.

And the oblique asymptotes are given by $f(x)=mx+n+\frac{1}{x}$. So the oblique asymptote in this case is $\\$ $y=x$.

So, the graph is

enter image description here

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Awesome explanation. –  moreno94 Mar 30 '12 at 10:14
    
@Garmen1778 I'm not sure if you were implying that there was a horizontal asymptote or not, but in case there's any ambiguity, there are no horizontal asymptotes. –  AlanH May 14 '13 at 17:46
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Try seeing the limit of the difference of this function and $f(x)=x$. Since $\dfrac{1}{x}\rightarrow 0$ when $x\rightarrow\pm\infty$, what can you tell since the limit of the difference is zero?

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