## Argos 2 years ago Pls respond

1. Argos

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2. RONNCC

ok so what do you think? :)

3. RONNCC

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. Argos

bamp

5. RONNCC

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

6. RONNCC

tell us what you find

7. Azteck

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

8. RONNCC

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

9. Argos

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

10. Argos

and can to use lhospitals rule on those but nega exponent

11. Azteck

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

12. Argos

13. Argos

because the back of the book sais its e

14. Argos

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. RolyPoly

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. binarymimic

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

17. shubhamsrg

shouldnt answer be 2 ? from binomial approximation ?

18. RolyPoly

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. RolyPoly

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

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24. RolyPoly

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$