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 :/

Headingnorth
May 9, 2015

#3**+10 **

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

Alan
May 9, 2015

#4**+5 **

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!

Headingnorth
May 9, 2015

#5**+15 **

Best Answer

I've changed it!

L'Hopital's rule:

limit_{x→infinity}(f(x)/g(x)) = limit_{x→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

.

Alan
May 9, 2015

#6**+5 **

Oops, thank you Alan.

I´m to slow ;)

Have a nice day and thanks for the rule!

Headingnorth
May 9, 2015