## anonymous 3 years ago Pls respond

1. anonymous

|dw:1357625465697:dw|

2. anonymous

ok so what do you think? :)

3. anonymous

the goal is to help, not to tell you the answer. sorry, we can't help on your test ;). So look up l'hopital and indeterminate form next.

4. anonymous

bamp

5. anonymous

yup. look up l'hopital and "indeterminate" form.

6. anonymous

tell us what you find

7. anonymous

Dude Hero gave it away man. 1/x vanishes and then you have $(1)^{-x}$

8. anonymous

@Azteck how did he give it away? . ok first get @Argos to find out what indeterminate form is -____-

9. anonymous

i know inderminite form is 0/0 or inf/inf

10. anonymous

and can to use lhospitals rule on those but nega exponent

11. anonymous

Then you get this: $\frac{ 1 }{ (1){^x} }$ $\frac{ 1 }{ (1)^{\infty} }$ $=\frac{ 1 }{ 1 }$

12. anonymous

13. anonymous

because the back of the book sais its e

14. anonymous

nope I even doubel checked on wolfram http://www.wolframalpha.com/input/?i=lim+as+x-%3E+inf+of+%281-%281%2Fx%29%29^-x

15. anonymous

Limit x approaches to infinity ( 1+ (1/x) )^x = e In your case, just replace x by -x in this formula. So, it's e.

16. anonymous

yeah 1 to the infinity is indeterminate so substitution will not work

17. shubhamsrg

shouldnt answer be 2 ? from binomial approximation ?

18. anonymous

Shouldn't it be e? A more accurate answer?

19. shubhamsrg

i can see easily why e fits here.. but i dont understand why doesnt the ans come from binomial approximation

20. anonymous

Sorry, I don't know binomial...

21. shubhamsrg

ahh,maybe i see why.. power should be a real no. for that, here, power is not a real no.

22. shubhamsrg

so, e is correct..

23. amoodarya

|dw:1357649297414:dw|

24. anonymous

Even though we use L'Hopital's Rule, we get the same answer.. Let $$y= (1-\frac{1}{x})^{-x}$$ $\ln y= -x \ln (1-\frac{1}{x})$$\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}-x \ln (1-\frac{1}{x})$$\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}\frac{ \ln (1-\frac{1}{x})}{-\frac{1}{x}}$Use L'Hopital's Rule:$\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}\frac{\frac{1}{{(1-\frac{1}{x})}}(\frac{1}{x^2})}{{\frac{1}{x^2}}}$$\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}\frac{1}{{(1-\frac{1}{x})}}$$\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}\frac{1}{{(\frac{x-1}{x})}}$$\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}\frac{x}{x-1}$Use L'Hopital's Rule again.$\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}\frac{1}{1}$$\lim_{x \rightarrow \infty}\ln y= 1$$\lim_{x \rightarrow \infty}e^{\ln y}= e^1$$\lim_{x \rightarrow \infty}y= e^1$$\lim_{x \rightarrow \infty}(1-\frac{1}{x})^{-x}= e$