Asked by Sean
Limit as n approaches infinity:
[(n + 3)/(n + 1)]^n
How do I even start this?
[(n + 3)/(n + 1)]^n
How do I even start this?
Answers
Answered by
bobpursley
I would divide the polynomials...
lim (1+ 2/n + ...)^n= 1
lim (1+ 2/n + ...)^n= 1
Answered by
Sean
Answer is e^2.
lim (n -> infinity) of [(n+3)(n+1)]^n
= lim (n -> infinity) of [1 + 2/(n+1)]^n
e = lim (x -> infinity) of [1 + 1/x]^x
e^2 = lim (x -> infinity) of [1 + 2/x + 1/x^2]^x
lim (n -> infinity) of [(n+3)(n+1)]^n
= lim (n -> infinity) of [1 + 2/(n+1)]^n
e = lim (x -> infinity) of [1 + 1/x]^x
e^2 = lim (x -> infinity) of [1 + 2/x + 1/x^2]^x
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