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]}