How do you identify all asymptotes or holes for #f(x)=(x^2-2)/x#?

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Answer 1 · 2017-01-18T07:58:11

The vertical asymptote is #x=0#
The slant asymptote is #y=x#
No holes and no horizontal asymptote.

The domain of #f(x)# is #D_f(x)=RR-{0}#

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

The vertical asymptote is #x=0#

We can rewrite #f(x)# as

#f(x)=(x^2-2)/x=x-2/x#

Therefore,

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

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

The slant asymptote is #y=x#

graph{(y-(x^2-2)/x)(y-x)=0 [-11.25, 11.25, -5.625, 5.62]}