Using L'Hospital's Rule to Evaluate a Limit

[1337caesar]

so I've been working on this for a while now and i just can't get it.

Evaluate the limit using L'Hospital's rule if necessary

lim X -> INF
(x/x+1) ^7x

i know i need to do...


exp(lim x-> inf 7x ln(x/x+1))

from there I'm stuck.

i appreciate any help

 

[lzkelley]

You could break it up into numerator and denominator, f(x)/g(x) let's say, where f(x) = x^7x and g(x) = (x+1)^7x l'hopital's rule says that if both f(x) and g(x) are divergent, then x--> inf f(x)/g(x) = x-->inf f'(x) / g'(x) , so you take the derivative of the top and bottom separately... you're doing to have to do this until its not divergent anymore... or 8 times? i guess?
i see what you're going for with the exp(7xln...) but i think that's probably harder / more work. Good luck

 

[Dick]

7x*ln(x/(x+1)) is a 0*infinity limit. Use l'Hopital again. Write it as 7*ln(x/(x+1))/(1/x). You can make life a little easier by changing ln(x/(x+1)) to -ln((x+1)/x) and simplifying inside the log.

 

[Dick]

well, we can see that the answer is going to be one, because in the limit that x --> inf, x = x+1 therefore x/x+1 = 1, so 1^anything even inf is still 1.
Now to prove that more rigorously, you break it up into numerator and denominator, f(x)/g(x) let's say, where f(x) = x^7x and g(x) = (x+1)^7x l'hopital's rule says that if both f(x) and g(x) are divergent, then x--> inf f(x)/g(x) = x-->inf f'(x) / g'(x) , so you take the derivative of the top and bottom separately... you're doing to have to do this until its not divergent anymore... or 8 times? i guess?
i see what you're going for with the exp(7xln...) but i think that's probably harder / more work. Good luck

The limit isn't 1. And taking the log does make it easier.

 

[exk]

As Dick said taking the log makes it a lot easier.

You may have to apply l'Hopital more than once.

 

[1337caesar]

thanks guys i got it by taking ln of the whole thing and l'hopitaling the thing twice. ended up being e^(-7)