anonymous
  • anonymous
Just a small question...
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Having something like this... \[\lim _{x \rightarrow0}e ^{(\frac{ 3 }{ x })\ln(1-3x)}\] is it possible to bring the e out and get this... \[e \lim _{x \rightarrow0}(\frac{ 3 }{ x })\ln (1-3x)\]
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \lim_{x\rightarrow 0}e^{\frac{3}{x}\ln(1-3x)} }\) \(\large\color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{3}{x}\ln(1-3x)} }\)
SolomonZelman
  • SolomonZelman
is this better ?

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anonymous
  • anonymous
Oh yes! That's it! But what's the logic behind this step? I don't exactly understand
SolomonZelman
  • SolomonZelman
you should review the limit properties
anonymous
  • anonymous
Hmmm ok gotcha!
SolomonZelman
  • SolomonZelman
I mean, nothing offensive, but find a link and read them over. you can find them absolutely everywhere
anonymous
  • anonymous
No worries at all! Thanks for the help!
SolomonZelman
  • SolomonZelman
\(\LARGE\color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{x}} }\) \(\LARGE\color{black}{ \displaystyle e^{-\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{-x}} }\)
SolomonZelman
  • SolomonZelman
did this step just now make sense?
SolomonZelman
  • SolomonZelman
now both top and bottom go to -infinity
SolomonZelman
  • SolomonZelman
go ahead and apply L'H's
anonymous
  • anonymous
Is that the same as what you wrote before? (e^ lim....)
SolomonZelman
  • SolomonZelman
I didn't change the value. I didn't do any unallowed step
SolomonZelman
  • SolomonZelman
am I allowed to multiply times -1 twice? (on the bottom and in front of the lim)
SolomonZelman
  • SolomonZelman
\(\LARGE\color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{x}} }\) \(\LARGE\color{black}{ \displaystyle e^{-\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{-x}} }\)
SolomonZelman
  • SolomonZelman
now apply the lhs to the limit. questions about how I got till there?
anonymous
  • anonymous
would we have to multiply by -1 ?
SolomonZelman
  • SolomonZelman
what do you mean? where?
SolomonZelman
  • SolomonZelman
I already got it to approach -oo
SolomonZelman
  • SolomonZelman
to approach -oo on top and bottom
anonymous
  • anonymous
Ohhh but it should be approaching 0
SolomonZelman
  • SolomonZelman
oh, I mean 0
SolomonZelman
  • SolomonZelman
it is.
SolomonZelman
  • SolomonZelman
ln(1-0)=1 -x=-0=0
SolomonZelman
  • SolomonZelman
why am I saying -oo ? apologize
SolomonZelman
  • SolomonZelman
\(\LARGE\color{black}{ \displaystyle e^{-\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{-x}} }\) top and bottom DO go to 0. ln(1)=0 and -(0)=0
anonymous
  • anonymous
So it's just the first line right?
anonymous
  • anonymous
I mean I don't get why we'd have a -lim and -x in the denominator
SolomonZelman
  • SolomonZelman
\(\LARGE\color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{x}} }\)
SolomonZelman
  • SolomonZelman
it was my mistake, i was thinking of infinity b/c i thought the lim -> (-oo)
SolomonZelman
  • SolomonZelman
but in can x->0 take LHS as it is
SolomonZelman
  • SolomonZelman
\(\LARGE\color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{x}} }\) \(\huge \color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{3\frac{-3}{1-3x}}{1}} }\)
anonymous
  • anonymous
Oh okie dokie! So now its just L'Hopital right?
SolomonZelman
  • SolomonZelman
yes... that i did d/dx on top and bottm
SolomonZelman
  • SolomonZelman
\(\huge \color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{-9}{1-3x}} }\)
anonymous
  • anonymous
I get -9 on the top but on the bottom isn't the derivative of x just 1?
SolomonZelman
  • SolomonZelman
yes, 3 replies ago i did the derivative
anonymous
  • anonymous
Ohh cause its a fraction over a fraction!
anonymous
  • anonymous
Yeah got it!!
SolomonZelman
  • SolomonZelman
wel, 1 on bottom for x, and the derivative of ln(1-3x) is a fraction -3/(1-3x)
SolomonZelman
  • SolomonZelman
ok, what is your answer?
anonymous
  • anonymous
is it e^-9 ? So 1/e^9
SolomonZelman
  • SolomonZelman
yes, there you go!
SolomonZelman
  • SolomonZelman
I can review the limit properties with you if you want...
anonymous
  • anonymous
Finally! Haha! Thanks so much! I've been bugging you a lot today! Im so sorry!
SolomonZelman
  • SolomonZelman
Ok, you welcome. You can choose to rvw limit properties later alone, or if you would like to, I can type more here...
anonymous
  • anonymous
That's fine! I can look it up on Google real quick! :D
anonymous
  • anonymous
Thanks again!
SolomonZelman
  • SolomonZelman
Alright. Enjoy:) ... not a problem.

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