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

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Answer 1 · 2016-11-20T07:55:24

The vertical asymptote is $x=0$
No slant asymptote
The horizontal asymptote is $y=0$

As you cannot divide by $0$
Domain, $D_g(x)=RR-{O}$

$x!=0$
Therefore, $x=0$ is a vertical asymptote

The degree of the numerator is $<$ the degree of the denominator.
So, we don't have a slant asymptote.

For the limits of $y$ , we take the terms of highest coefficients

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

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

Therefore, $y=0$ is a horizontal asymptote

graph{(x+2)/(2x^2) [-7.02, 7.03, -3.51, 3.51]}

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