Indeterminate? lim [t->infty] [-(t 2^(-t)) / ln(2)]

[T_TEngineer_AdamT_T]

\(\displaystyle \L \lim_{t\to +\infty} -\frac{t2^{-t}}{\ln{2}}\)
is this just equal to zero? or will i still evaluate it using LHR

i think so
\(\displaystyle \L -\frac{\infty}{2^{\infty}\ln2}\)

using LHR:
\(\displaystyle \L \lim_{t\to +\infty} -\frac{t}{2^{t}\ln{2}}\)
\(\displaystyle \L \lim_{t\to +\infty} -\frac{1}{2^{t}(\ln2)^2}\)
\(\displaystyle \L = 0\)

 

[galactus]

You're right, Adam. It's 0