Question

i need help finding the limit of the sequence

(1 + 1/n) ^ n using l'hopitals rule

Answer 1

the limit as n goes to infinity I assume?

Answer 2

yes sir

Answer 3

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

Answer 4

you almost did 75% of the problem

Answer 5

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

Answer 6

so the denominator would be -1/x^2

Answer 7

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

Answer 8

which is equal to 1 as you can see.

Answer 9

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

Answer 10

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

Answer 11

yes you're right!!

Answer 12

awesome thanks bud

Answer 13

no problem.. just be a fan ;)

Answer 14

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

Answer 15

x is approaching what?

Answer 16

infinty again

Answer 17

use the same procedure we did with the first one.

Answer 18

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

Answer 19

you should use parentheses.