Question

horizontal asymptote of

Answer 1

HI!!

Answer 2

this would depend on if \(x\to \infty\) or if \(x\to -\infty\)

Answer 3

in other words it has two horizontal asympotes

Answer 4

the denominator is always positive, but the numerator would be positive as \(x\to \infty\), but negative as \(x\to -\infty\)

Answer 5

hope it is pretty clear that it is either \(y=1\) or \(y=-1\) depending on which way you go

Answer 6

\[\frac{x}{\sqrt{x^2+1}}=\frac{x}{\sqrt{x^2(1+\frac{1}{x^2})}}\]
\[=\frac{x}{\sqrt{x^2}\sqrt{1+\frac{1}{x^2}}}=\frac{x}{|x|\sqrt{1+\frac{1}{x^2}}}\]
\[=sgn(x)\frac{1}{\sqrt{1+\frac{1}{x^2}}}\]

Answer 7

but either way isnt an infinty divided by infinty infinity? @misty1212