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

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

Mathematics
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log is continuous function, so you can send the limit inside the function : \[\lim(\log(f(x))) = \log(\lim f(x))\]
\[\lim_{x \rightarrow \infty}\left[ \log_{5}(\frac{ 1 }{ 125 }-2^{-x}) \right] = \log_5\left[ \color{blue}{\lim_{x \rightarrow \infty}(\frac{ 1 }{ 125 }-2^{-x})} \right]\]
oh, okay

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What do I do next?
you can do many things maybe just think of what happens to \(\large 2^{-x}\) as you make \(x\) large
\(2^{-1} = ?\) \(2^{-2} = ?\) \(2^{-3} = ?\) \(\cdots\) \(2^{-100} = ?\)
Doesn't the value become smaller?
evaluate those values and see
.5 .25 .125 7.8888...E-31
you can see the value of \(2^{-x}\) is approaching \(0\) as you increase \(x\) so \[\lim\limits_{x\to\infty}2^{-x} = 0\]
\[\begin{align} \lim_{x \rightarrow \infty}\left[ \log_{5}(\frac{ 1 }{ 125 }-2^{-x}) \right] &= \log_5\left[ \color{blue}{\lim_{x \rightarrow \infty}(\frac{ 1 }{ 125 }-2^{-x})} \right]\\~\\ &=\log_5\left[ \color{blue}{\frac{ 1 }{ 125 }-0} \right]\\~\\ &=\log_5\left[ \color{blue}{5^{-3}} \right]\\~\\ &=-3 \end{align}\]
oohh! I see. Thanks for explaining this. Greatly appreciated! :) I'm not really good at limits and thinking about infinities and such. Any tips?
my only tip is not to try and visualize everything, sometimes you need to just follow the rules and things will be simple
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
Oh, okay. I will keep that in mind. Thanks for everyting! :)

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