MAT334--2020F > Quiz 2

Quiz2 Lec5101 5D

(1/1)

Yuyan Liu:
Question: Find the limit at $\infty$ of the given function, or explain why it does not exist:
$g(z) = \frac {4 z^6 - 7 z^3}{(z^2 - 4)^3}$
Answer: $g(z) = \frac {4 z^6 - 7 z^3}{z^6 + 3(-4) z^4 + 3(-4)^2  z^2 + (-4)^3} = \frac {4 z^6 - 7 z^3}{z^6 - 12 z^4 + 48 z^2 - 64}$
Divide both denominator and numerator by $z^6$:
$g(z) = \frac {4 - 7 z^{-3}}{1 - 12 z^{-2} + 48 z^{-4} - 64 z^{-6}}$
$$ \lim_{z \to\infty} g(z) = \frac {4 - 7 \infty ^{3}}{1 - 12 \infty ^{-2} + 48 \infty ^{-4} - 64 \infty ^{-6}} = \frac {4 - 0}{1 - 0 +0 - 0} = 4$$

Navigation

[0] Message Index