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How do you find the limit of #(x/(x+1))^x# as x approaches #oo#?

Answer 1

Emilia Aguilar

The limit is #l=e^-1=1/e#

we can write this as

#e^[x*ln(x/(1+x))]#

Hence actually we need to find the limit

#lim_(x->oo) [x*ln(x/(1+x))]=lim_(x->oo) [ln(x/(1+x))/(1/x)]#

Because

#ln(x/(1+x))/[1/x]->0/0# apply L'Hopital law hence

#([ln(x/(1+x))]')/([1/x]')=[1/(x^2+x)]/(-1/x^2)= -x^2/(x^2+x)=-x/(1+x)->-1# as #x->oo#

Hence the limit is #e^-1=1/e#

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Answer 2

Ezra Adams

To find the limit of (x/(x+1))^x as x approaches infinity, we can use the concept of exponential limits.

First, we rewrite the expression as (1 + 1/x)^(-x).

Next, we take the limit as x approaches infinity.

Using the limit property, we have e^(-1) as the limit, where e is the base of the natural logarithm.

Therefore, the limit of (x/(x+1))^x as x approaches infinity is e^(-1).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.