Question

lim [(4^n + 5^n)^1/n] = ?
n->infinity

Answer 1

I'm thinking logarithmic might be the way to go here... not sure yet.

Answer 2

I took the logarithm. After that should I proceed with L Hospital rule? Although I did it but didn't arrive to the result. Please help

Answer 3

What did you get when using l'hopital?

Answer 4

=lim[n->infty] e^{[ln(4^n+5^n)]/n}
=e^{lim[n->infty][ln(4^n+5^n)]/n}
=e^{lim[n->infty](4^nln4+5^nln5)/(4^n+5^n)}

Answer 5

yes, I got this exctly. But how to proceed further?

Answer 6

Well.......

Answer 7

Any people here know?

Answer 8

yikes... that looks nasty :P

Answer 9

lemme see, I think I might have something...

Answer 10

\[\lim_{n\rightarrow \infty} (4^n + 5^n)^{1/n} = \lim_{n\rightarrow \infty}(5^n[(\frac{4}{5})^n + 1^n])^{1/n} = \lim_{n \rightarrow \infty}5[(\frac{4}{5})^n + 1]^{1/n} = 5 \lim_{n\rightarrow \infty}[(\frac{4}{5})^n + 1]^{1/n}\]

Answer 11

Is that a fail

Answer 12

so now, we use ln on the remaining limit portion

Answer 13

\[\lim_{n \rightarrow \infty} \frac{1}{n}\ln(1 + [4/5]^n)=0\] because ln(1) = 0.

Answer 14

This gives the final answer of
\[5 e^{limit thing} = 5e^0 = 5\]

Answer 15

OK, got that clearly. Thank you.

Answer 16

no problem, that was tricky :P

Answer 17

Wow