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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

Best Answer 

 #5
avatar+26322 
+15

I've changed it!

 

L'Hopital's rule:

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

.

Alan  May 9, 2015
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6+0 Answers

 #1
avatar+90988 
+5

Is that a 6x on the bottom Headingnorth?

I probably can't do it anyway.  :(

Melody  May 9, 2015
 #2
avatar+223 
+5

oh sorry it will be 3*2^x

Headingnorth  May 9, 2015
 #3
avatar+26322 
+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
avatar+223 
+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
avatar+26322 
+15
Best Answer

I've changed it!

 

L'Hopital's rule:

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

.

Alan  May 9, 2015
 #6
avatar+223 
+5

Oops, thank you Alan.

I´m to slow ;)

Have a nice day and thanks for the rule!

Headingnorth  May 9, 2015

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