Toronto Math Forum
MAT3342018F => MAT334Tests => Quiz2 => Topic started by: Victor Ivrii on October 05, 2018, 06:11:52 PM

Find the limit as $z\to \infty$ of the given function, or explain why it
does not exist:
\begin{equation*}
g(z)=\frac{4z^6 7 z^3}{(z^24)^3}.
\end{equation*}

$$g(z) = \frac{4z^67z^3}{(Z^24)^3} = \frac{4z^67z^3}{z^612z^4+48z^264}$$
we can divide $z^6$ on both numerator and denominator.
Then we can get
$$g(z) = \frac{47z^{1}}{z12z^{2}+48z^{4}64z^{6}}$$
Then as $z\to\infty $
$$\lim_{z\to\infty} f(z) = \frac{4  0}{1  0 + 0  0} = 4$$