Question

Do the problem in the attached file

Answer 1

This problem is from my site http://mathonline.missouri.edu

Answer 2

\[\lim_{t\to\infty}(1+\frac5t+\frac{10}{t^4})^{9t}=\exp(\lim_{t\to\infty}9t\ln(1+\frac5t+\frac{10}{t^4})\]\[=\exp(\lim_{n\to0}{9\ln(1+5n+10n^4)\over n})\]now use l'Hospital...

Answer 3

\[=\exp(\lim_{n\to0}{45+360n^3\over1+5n+10n^4})=\exp(45)=e^{45}\]

Answer 4

agrees with @TuringTest the answer must be e^45

Answer 5

@TuringTest i quite don't understand the third step in your fist post though

Answer 6

I made the substitution\[t=\frac1n\] so as \(t\to\infty\) that implies that \(n\to0\), which allowed me to use l'Hospital's rule

Answer 7

Oo... i never thought such elegant method existed. I had to assume (5/t+10/t^4) < 1 and expand with log series and cancel out t to get 45 though

Answer 8

Yeah, I learned that technique here on OS
it's more than just a way to practice, we can learn here too :)

Answer 9

I doubt I could even do your method though

Answer 10

it's like a brute force hack ,,, always gives an advantage to a thick minded person like me
log(1+x) = x+((x^2)/2)+((x^3)/3)+.............((x^n)/n)+...........

Answer 11

oh yeah, you're so dim
(*sarcastic smirk)
What do you study anyway?

Answer 12

Physics major

Answer 13

You??

Answer 14

Nice :)
Me too, but only aspiring at the moment
I'm a little older, so I don't know what I can do getting into school at my age

I gotta bounce for now, see ya 'round!

Answer 15

see ya