How do you find the Vertical, Horizontal, and Oblique Asymptote given $f (x) = \frac{x}{x^{2} + 4}$ $f(x)=x/(x^2+4)$ ?

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Answer 1 · 2017-01-28T06:26:13

The horizontal asymptote is $y = 0$ $y=0$
No vertical asymptote
No oblique asymptote

The domain of $f (x)$ $f(x)$ is $D_f(x)=RR$

The denominator is $>0 AA x in RR$

There are no vertical asymptotes.

As the degree of the numerator is $<$ than the degree of the denominator, there is no oblique asymptote.

$lim_(x->-oo)f(x)=lim_(x->-oo)x/x^2=lim_(x->-oo)1/x=0^-$

$lim_(x->+oo)f(x)=lim_(x->+oo)x/x^2=lim_(x->+oo)1/x=0^+$

The horizontal asymptote is $y=0$

graph{(y-x/(x^2+4))(y)=0 [-5.55, 5.55, -2.773, 2.776]}

How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x)=x/(x^2+4)f(x)=xx2+4f(x)=x/(x^2+4)?