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!