Question

Why does the limit of nln(1+1/n) as n approaches infinity = 1? Wouldn't the expression in the brackets become very close to 1, essentially turning the expression into nln(1) = 0?

Answer 1

suppose f(x)=n and g(x)=ln(1+1/n), then lim fg= lim f * lim g only if both of the limits exist. since limit of f as n goes to infinity does not exist, then you can't apply that. So you cant take the limit of ln(1+1/n) to be 0 and say that 0 times anything is 0 so that is your answer.



there are multiple ways of doing this, but i'll do it with a substitution to put it in a form that i can then use lhopitals rule.
let h=1/n
then lim h as n goes to infinity is 0

so now the equation looks like lim h->0 [ln(1+h)]/h

this is the indeterminate form 0/0, so we can apply lhopitals rule.

lim f/g = lim f'/g'
so now it is lim h->0 [(1/1+h)]/1

which then just ends up as 1, so lhopitals rule is valid, and there is your limit.