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Consider the function $z(x)=\sqrt{1+x^2}+1$

Show that $y=x+1$ and $y=1-x$ are linear asymptotes of the function at $\infty$ and respectively $- \infty$

So I started of with the first part: show that $y=x+1$ is a linear asymptote of the function at $\infty$:

Note that $\forall x > 0$:

$|z(x)-(1+x)|=|1+\sqrt{1+x^2}-1-x|=|\sqrt{1+x^2}-x|=\sqrt{1+x^2}-x$

But if I go on extracting a value for x with the use of $\epsilon$, I find something I don't consider true, since i find $x>\frac{1-\epsilon^2}{2\epsilon}$

Am I doing wrong, or can somebody tell me what I missed out? Note that I only have to show, not proof.

Many thanks!

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    If f is a function on the interval $(a,\infty)$, there's an asymptote <==> f(x) converges to $mx+b$ when $x \to \infty$, if \forall \epsilon > 0 there exists a number Q such that |f(x)-(mx+b)|<\epsilon whenever x>Q2012-11-06

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What you want to show is that $\lim_{x\to\infty}\left(\sqrt{1+x^2}-x\right)=0\;.$ Rewrite the function:

$\sqrt{1+x^2}-x=\left(\sqrt{1+x^2}-x\right)\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}+x}=\frac1{\sqrt{1+x^2}+x}\;.$

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    @Bob: You’re welcome, and thank *you*.2012-11-06