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

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Answer 1 · 2017-01-09T06:36:48

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

The domain of $g(x)$ is $D_g(x)=RR-{0}$

As we cannot divide by $0$ , $x!=0$

The vertical asymptote is $x=0$

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

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

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

The horizontal asymptote is $y=0$

graph{(y-(x+2)/(2x^2))(y)(y-100x)=0 [-7.02, 7.024, -3.51, 3.51]}

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