find the limit:

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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|dw:1444612484256:dw|
what do i do with the 3x?
because i know there is formula that says e^1 if there was only an x in the exponent instead of a 3x

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Hmm ya, it's very similar to the definition of e, huh? :) You can either use logarithms to get this beast done, or you can relate it back to the definition of e.
\[\large\rm \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{3x}=\quad \lim_{x\to\infty}\left[\left(1+\frac{1}{x}\right)^x\right]^{3}=\quad \left[\color{orangered}{\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x}\right]^{3}\]
\[\large\rm =\left[\color{orangered}{e}\right]^3\]So ya that's one clever way to do it. Your teacher would probably prefer you use logs or L'Hopital or something though. Hmm
Do you understand what I did with the exponent? I was applying exponent rule in reverse: \(\large\rm a^{xy}=(a^x)^y\)
yes that completely makes sense! thanks! btw how would i use logs then?
Recall that the exponential and log are inverse functions of one another. So when you take their composition, you just get the argument back.\[\large\rm e^{\ln(\color{orangered}{x})}=\color{orangered}{x}\]We can use this idea to get the 3x out of the exponent. This might seem a little strange, try to follow the color though.\[\large\rm \color{orangered}{\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{3x}}=e^{\ln\left[\color{orangered}{\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{3x}}\right]}\]Let's pass the limit outside of the log, and ignore the exponential part of it, just for now,\[\large\rm \lim_{x\to\infty}\ln\left[\left(1+\frac{1}{x}\right)^{3x}\right]\]
From here, your log rule lets you pull the 3x out,\[\large\rm \lim_{x\to\infty}~3x\cdot\ln\left(1+\frac{1}{x}\right)\]Then you have to rewrite your 3x in a clever way, rewrite it as 1/(1/3x)\[\large\rm \lim_{x\to\infty}\frac{\ln\left(1+\frac{1}{x}\right)}{\frac{1}{3x}}\]It's one of those weird tricks you have to remember I guess... And from here, this whole mess is approaching the indeterminate form 0/0 as x approaches infinity. So you can apply L'Hopital's rule.
This is definitely not the most fun approach :) lolol lots of little tricky steps.
okay! Thank You! :)
So anyway, you apply L'Hopital, you can evaluate your limit successfully after that, it should give you 3. So then you back up to that exponential that we left out earlier. It was e^(lim stuff) and we figured out that it's e^(3). Ya, just something to think about :P

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