Question

Evaluate.
\[\lim_{x \rightarrow \infty}\left[ \log_{5}(\frac{ 1 }{ 125 }-2^{-x}) \right]\]

Answer 1

log is continuous function, so you can send the limit inside the function :
\[\lim(\log(f(x))) = \log(\lim f(x))\]

Answer 2

\[\lim_{x \rightarrow \infty}\left[ \log_{5}(\frac{ 1 }{ 125 }-2^{-x}) \right] = \log_5\left[ \color{blue}{\lim_{x \rightarrow \infty}(\frac{ 1 }{ 125 }-2^{-x})} \right]\]

Answer 3

oh, okay

Answer 4

What do I do next?

Answer 5

you can do many things
maybe just think of what happens to \(\large 2^{-x}\) as you make \(x\) large

Answer 6

\(2^{-1} = ?\)
\(2^{-2} = ?\)
\(2^{-3} = ?\)
\(\cdots\)
\(2^{-100} = ?\)

Answer 7

Doesn't the value become smaller?

Answer 8

evaluate those values and see

Answer 9

.5
.25
.125
7.8888...E-31

Answer 10

you can see the value of \(2^{-x}\) is approaching \(0\) as you increase \(x\)
so
\[\lim\limits_{x\to\infty}2^{-x} = 0\]

Answer 11

\[\begin{align}
\lim_{x \rightarrow \infty}\left[ \log_{5}(\frac{ 1 }{ 125 }-2^{-x}) \right] &= \log_5\left[ \color{blue}{\lim_{x \rightarrow \infty}(\frac{ 1 }{ 125 }-2^{-x})} \right]\\~\\
&=\log_5\left[ \color{blue}{\frac{ 1 }{ 125 }-0} \right]\\~\\
&=\log_5\left[ \color{blue}{5^{-3}} \right]\\~\\
&=-3
\end{align}\]

Answer 12

oohh! I see. Thanks for explaining this. Greatly appreciated! :) I'm not really good at limits and thinking about infinities and such. Any tips?

Answer 13

my only tip is not to try and visualize everything, sometimes you need to just follow the rules and things will be simple

Answer 14

not meant to say, stop visualizing... just want to say that following rules is also important as calculus is very huge, learning wont be smooth w/o a systematic approach

graph everything but don't always try to understand in terms of graphs only
https://www.desmos.com/calculator

Answer 15

Oh, okay. I will keep that in mind. Thanks for everyting! :)