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## anonymous 5 years ago i need help finding the limit of the sequence (1 + 1/n) ^ n using l'hopitals rule

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1. anonymous

the limit as n goes to infinity I assume?

2. anonymous

yes sir

3. anonymous

I understand taking the ln of it so you get n ln (1+1/n) then you have ln(1+1/n) / 1/n but then i am lost

4. anonymous

you almost did 75% of the problem

5. anonymous

now you're ready to apply the l'hopital's rule, that's taking the derivative of both The numerator and denominator

6. anonymous

so the denominator would be -1/x^2

7. anonymous

you will get: $\lim_{n \rightarrow \infty} {\ln (1+1/n) \over 1/n}=\lim_{n \rightarrow \infty} { {-1/n^2 \over 1+1/n} \over -1/n^2}=\lim_{n \rightarrow \infty}{1 \over 1+1/n}$

8. anonymous

which is equal to 1 as you can see.

9. anonymous

so the top is 1 / 1 + 1/n and then you take the derivate of the inside of ln (1 + 1/n) right

10. anonymous

but this is the limit of the function after taking the ln, so the limit of your original function is e^1

11. anonymous

yes you're right!!

12. anonymous

awesome thanks bud

13. anonymous

no problem.. just be a fan ;)

14. anonymous

AnwarA you still there I got another tought one its (1 + 1/sqrt(x))^x

15. anonymous

x is approaching what?

16. anonymous

infinty again

17. anonymous

use the same procedure we did with the first one.

18. anonymous

ok so I got to that part I got ( -1/2x ^ 3/2)/ 1 + 1/sqrt(x) / -1/x^2 but it doesnt look good to simplify to me

19. anonymous

you should use parentheses.

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