Question

How do you find the limit as n goes to infinity of ( n^6 + 2^(n-1) )/((2/n) +(2^(n+1)))

Answer 1

i can say it in english, but not sure of what method you could use. maybe l'hopital
in english it is easy. the only terms that matter are
\[2^{n-1}\]a nd
\[2^{n+1}\] because the 6/n goes to zero and n^6 is negligible compared to the exponential part

Answer 2

and
\[\frac{2^{n-1}}{2^{n+1}}=\frac{1}{4}\] and that is your limit

Answer 3

Wait so are you saying that the first term doesn't matter? If so how come?