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amyna
 one year ago
find the limit:
amyna
 one year ago
find the limit:

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amyna
 one year ago
Best ResponseYou've already chosen the best response.0what do i do with the 3x?

amyna
 one year ago
Best ResponseYou've already chosen the best response.0because i know there is formula that says e^1 if there was only an x in the exponent instead of a 3x

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Hmm 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{3x}=\quad \lim_{x\to\infty}\left[\left(1+\frac{1}{x}\right)^x\right]^{3}=\quad \left[\color{orangered}{\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x}\right]^{3}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm =\left[\color{orangered}{e}\right]^3\]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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Do you understand what I did with the exponent? I was applying exponent rule in reverse: \(\large\rm a^{xy}=(a^x)^y\)

amyna
 one year ago
Best ResponseYou've already chosen the best response.0yes that completely makes sense! thanks! btw how would i use logs then?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Recall 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,\[\large\rm \lim_{x\to\infty}\ln\left[\left(1+\frac{1}{x}\right)^{3x}\right]\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2From 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2This is definitely not the most fun approach :) lolol lots of little tricky steps.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So 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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