I am trying to find the limit of the sequence
(3−4n1+n)(1+1n)n
I am aware that if one sequence converges and another sequence converges then the multiplication of two sequences also converge. The limit of the first sequence is −4. However I do not know how to calculate the limit of the second sequence.
Answer
Here is one approach
(1+1n)n=enln(1+1n)=en(1n−12n2+…)=e1−12n+…⟶∞e
No comments:
Post a Comment