How do you find vertical, horizontal and oblique asymptotes for $\frac{x^{2} - 16}{x^{4}}$ $(x^2-16)/(x^4)$ ?

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Answer 1 · 2016-10-29T19:23:47

The vertical asymptote is $x = 0$ $x=0$
The horizontal asymptote is $y = 0$ $y=0$

As you cannot divide by 0, so $x = 0$ $x=0$ is a vertical asymptote

The degree of the numerator is $>$ $>$ degree of the denominator, there is no oblique asymptote.

Limit $\frac{x^{2}}{x^{4}} = \frac{1}{x^{2}} = 0^{+}$ $x^2/x^4=1/x^2=0^+$
$x \to \pm \infty$ $x->+-oo$

So $y = 0$ $y=0$ is a horizontal asymptote

graph{(x^2-16)/x^4 [-6.93, 8.874, -5.09, 2.81]}

How do you find vertical, horizontal and oblique asymptotes for (x^2-16)/(x^4)x2−16x4(x^2-16)/(x^4)?