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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 $\forall x > 0$:


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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What is your definition of linear asymptote? – JavaMan Nov 6 '12 at 23:07
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>Q – Bob Nov 6 '12 at 23:14
up vote 3 down vote accepted

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


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Oh damn, I did think about the thing with multiplying the root out in the numerator.... But can you explain why we want to show your "to show" part? Since in my reader they dont talk about this way to show, they only show the $\epsilon$ "proof"..? – Bob Nov 6 '12 at 23:29
What they’re giving you is the $\epsilon$ proof that the limit is $0$: $\lim_{x\to\infty}\left(\sqrt{1+x^2}-x\right)=0$ just means that for every $\epsilon>0$ there is an $M\in\Bbb R$ such that $\left|\sqrt{1+x^2}-x\right|<\epsilon$ whenever $x\ge M$. – Brian M. Scott Nov 6 '12 at 23:36
Oh thats good to know, useful post and answer! Thanks :) – Bob Nov 6 '12 at 23:38
@Bob: You’re welcome, and thank you. – Brian M. Scott Nov 6 '12 at 23:41

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