- #1

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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

- Thread starter 1337caesar
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- #1

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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

- #2

- 275

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You could break it up into numerator and denominator, f(x)/g(x) lets 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 thats probably harder / more work. Good luck

i see what you're going for with the exp(7xln...) but i think thats probably harder / more work. Good luck

Last edited:

- #3

Dick

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Dick

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The limit isn't 1. And taking the log does make it easier.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) lets 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 thats probably harder / more work. Good luck

- #5

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

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

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

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