## anonymous one year ago Evaluate. $\lim_{x \rightarrow \infty}\left[ \log_{5}(\frac{ 1 }{ 125 }-2^{-x}) \right]$

1. rational

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

2. rational

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

3. anonymous

oh, okay

4. anonymous

What do I do next?

5. rational

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

6. rational

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

7. anonymous

Doesn't the value become smaller?

8. rational

evaluate those values and see

9. anonymous

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

10. rational

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

11. rational

\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}

12. anonymous

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

13. rational

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

14. rational

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

15. anonymous

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