Take limits to infinity. The power of x {\displaystyle x} in the denominator is greater than in the numerator and so the limits will both be zero.
lim x → ∞ x − 4 x ( x 2 − 16 ) = lim x → ∞ x − 4 x 3 − 16 x = lim x → ∞ 1 x 2 − 4 x 3 1 − 16 x 2 = 0 1 = 0. {\displaystyle \lim _{x\to \infty }{\frac {x-4}{x(x^{2}-16)}}=\lim _{x\to \infty }{\frac {x-4}{x^{3}-16x}}=\lim _{x\to \infty }{\frac {{\frac {1}{x^{2}}}-{\frac {4}{x^{3}}}}{1-{\frac {16}{x^{2}}}}}={\frac {0}{1}}=0.} We also have that: lim x → − ∞ x − 4 x ( x 2 − 16 ) = lim x → − ∞ x − 4 x 3 − 16 x = lim x → − ∞ 1 x 2 − 4 x 3 − 1 − 16 x 2 = 0 − 1 = 0. {\displaystyle \lim _{x\to -\infty }{\frac {x-4}{x(x^{2}-16)}}=\lim _{x\to -\infty }{\frac {x-4}{x^{3}-16x}}=\lim _{x\to -\infty }{\frac {{\frac {1}{x^{2}}}-{\frac {4}{x^{3}}}}{-1-{\frac {16}{x^{2}}}}}={\frac {0}{-1}}=0.}
in the denominator is greater than in the numerator and so the limits will both be zero.
We also have that:
