What is the limit of #(1+1/x)^x# as x approaches infinity?

Answer 1

#e#

We are trying to find:

#lim_(x->oo)[(1+1/x)^x]#

Let's begin by expanding the term in brackets using the binomial theorem

#(1+1/x)^x=1+((x),(1))1/x+((x),(2))1/x^2 +...+((x), (x))1/x^x #
such that the #k^(th)# term in the series is
#((x), (k))1/x^k=(x!)/(k!(x-k)!)*1/x^k#
We can simplify this expression by cancelling terms with #x# in them and gathering them together. Then we take the limit of each term in the sum
#lim_(x->oo)[1/(k!)*(x(x-1)(x-2)...(x-k+1))/(x^k)]#
From this we can see that the term on the right goes to 1 as #x->oo#. So the sum becomes:
#1/(0!)+1/(1!)+1/(2!) + ... #
This is the known series for the natural number, #e#, therefore
#lim_(x->oo)[(1+1/x)^x] = e#
Interestingly, this limit sometimes used to define #e#. For more information, see the "Series for e" section at the following link: https://tutor.hix.ai
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Answer 2

The limit of (1+1/x)^x as x approaches infinity is e, where e is the mathematical constant approximately equal to 2.71828.

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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.

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.

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.

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.

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