lim(x goes to infinity) (2^x+ln(x))/(3*2^x-ln(x))
Hope you guys can help me out with this one. I got stuck somewhere in the middle :/
Is that a 6x on the bottom Headingnorth?
I probably can't do it anyway. :(
2^x increases much quicker than ln(x) as x gets large so the numerator tends to 2^x as x gets very large.
3*2^x also increases much quicker than ln(x) as x gets large, so the denominator tends to 3*2^x
The ratio therefore tends to 2^x/(3*2^x) = 1/3
Thanks Melody for notice my wrong.
Alan, do you know how to do it with the change?
the answer is 1/3 if that helps.
Will check out L´Hopital´s rule, thanks for your help!
I've changed it!
limitx→infinity(f(x)/g(x)) = limitx→infinity(f`(x))/g`(x))
Here f`(x) = ln(2)*2^x + 1/x and g`(x) = 3*ln(2)*2^x - 1/x
as x gets large 1/x gets small so you are left with ln(2)*2^x/(3*ln(2)*2^x ) = 1/3