Question

Just a small question...

Answer 1

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... \[e \lim _{x \rightarrow0}(\frac{ 3 }{ x })\ln (1-3x)\]

Answer 2

\(\large\color{black}{ \displaystyle \lim_{x\rightarrow 0}e^{\frac{3}{x}\ln(1-3x)} }\)

\(\large\color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{3}{x}\ln(1-3x)} }\)

Answer 3

is this better ?

Answer 4

Oh yes! That's it! But what's the logic behind this step? I don't exactly understand

Answer 5

you should review the limit properties

Answer 6

Hmmm ok gotcha!

Answer 7

I mean, nothing offensive, but find a link and read them over. you can find them absolutely everywhere

Answer 8

No worries at all! Thanks for the help!

Answer 9

\(\LARGE\color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{x}} }\)

\(\LARGE\color{black}{ \displaystyle e^{-\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{-x}} }\)

Answer 10

did this step just now make sense?

Answer 11

now both top and bottom go to -infinity

Answer 12

go ahead and apply L'H's

Answer 13

Is that the same as what you wrote before? (e^ lim....)

Answer 14

I didn't change the value. I didn't do any unallowed step

Answer 15

am I allowed to multiply times -1 twice? (on the bottom and in front of the lim)

Answer 16

\(\LARGE\color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{x}} }\)

\(\LARGE\color{black}{ \displaystyle e^{-\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{-x}} }\)

Answer 17

now apply the lhs to the limit. questions about how I got till there?

Answer 18

would we have to multiply by -1 ?

Answer 19

what do you mean? where?

Answer 20

I already got it to approach -oo

Answer 21

to approach -oo on top and bottom

Answer 22

Ohhh but it should be approaching 0

Answer 23

oh, I mean 0

Answer 24

it is.

Answer 25

ln(1-0)=1
-x=-0=0

Answer 26

why am I saying -oo ? apologize

Answer 27

\(\LARGE\color{black}{ \displaystyle e^{-\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{-x}} }\)

top and bottom DO go to 0.

ln(1)=0
and -(0)=0

Answer 28

So it's just the first line right?

Answer 29

I mean I don't get why we'd have a -lim and -x in the denominator

Answer 30

\(\LARGE\color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{x}} }\)

Answer 31

it was my mistake, i was thinking of infinity b/c i thought the lim -> (-oo)

Answer 32

but in can x->0 take LHS as it is

Answer 33

\(\LARGE\color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{x}} }\)

\(\huge \color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{3\frac{-3}{1-3x}}{1}} }\)

Answer 34

Oh okie dokie! So now its just L'Hopital right?

Answer 35

yes... that i did d/dx on top and bottm

Answer 36

\(\huge \color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{-9}{1-3x}} }\)

Answer 37

I get -9 on the top but on the bottom isn't the derivative of x just 1?

Answer 38

yes, 3 replies ago i did the derivative

Answer 39

Ohh cause its a fraction over a fraction!

Answer 40

Yeah got it!!

Answer 41

wel, 1 on bottom for x, and the derivative of ln(1-3x) is a fraction -3/(1-3x)

Answer 42

ok, what is your answer?

Answer 43

is it e^-9 ? So 1/e^9

Answer 44

yes, there you go!

Answer 45

I can review the limit properties with you if you want...

Answer 46

Finally! Haha! Thanks so much! I've been bugging you a lot today! Im so sorry!

Answer 47

Ok, you welcome. You can choose to rvw limit properties later alone, or if you would like to, I can type more here...

Answer 48

That's fine! I can look it up on Google real quick! :D

Answer 49

Thanks again!

Answer 50

Alright. Enjoy:) ... not a problem.