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

Question

1061 views around the world

You can reuse this answer
Creative Commons License

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]}