How do you find the asymptotes for $f(x) = x / sqrt(x^2 + 1)$ ?

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Answer 1 · 2016-06-08T16:01:42

We can rewrite this as follows

$f(x)=(x)/[absxsqrt(1+1/x^2)]$

Hence for $x->+oo$ we have that $f(x)->1$

and

for $x->-oo$ we have that $f(x)->-1$

Hence the horizontal asympotes are $y=1$ and $y=-1$

How do you find the asymptotes for f(x) = x / sqrt(x^2 + 1)f(x) = x / sqrt(x^2 + 1)?