## anonymous one year ago Just a small question...

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

2. SolomonZelman

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

3. SolomonZelman

is this better ?

4. anonymous

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

5. SolomonZelman

you should review the limit properties

6. anonymous

Hmmm ok gotcha!

7. SolomonZelman

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

8. anonymous

No worries at all! Thanks for the help!

9. SolomonZelman

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

10. SolomonZelman

did this step just now make sense?

11. SolomonZelman

now both top and bottom go to -infinity

12. SolomonZelman

13. anonymous

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

14. SolomonZelman

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

15. SolomonZelman

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

16. SolomonZelman

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

17. SolomonZelman

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

18. anonymous

would we have to multiply by -1 ?

19. SolomonZelman

what do you mean? where?

20. SolomonZelman

I already got it to approach -oo

21. SolomonZelman

to approach -oo on top and bottom

22. anonymous

Ohhh but it should be approaching 0

23. SolomonZelman

oh, I mean 0

24. SolomonZelman

it is.

25. SolomonZelman

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

26. SolomonZelman

why am I saying -oo ? apologize

27. 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

28. anonymous

So it's just the first line right?

29. anonymous

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

30. SolomonZelman

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

31. SolomonZelman

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

32. SolomonZelman

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

33. SolomonZelman

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

34. anonymous

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

35. SolomonZelman

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

36. SolomonZelman

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

37. anonymous

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

38. SolomonZelman

yes, 3 replies ago i did the derivative

39. anonymous

Ohh cause its a fraction over a fraction!

40. anonymous

Yeah got it!!

41. SolomonZelman

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

42. SolomonZelman

43. anonymous

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

44. SolomonZelman

yes, there you go!

45. SolomonZelman

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

46. anonymous

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

47. SolomonZelman

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

48. anonymous

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

49. anonymous

Thanks again!

50. SolomonZelman

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