Lim x-->∞ (1+

^{1}/_{x})^{x}Lim x-->∞ (1+^{1}/_{x})^{x}

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New Wilmington, PA

Hi David,

The limit lim _{x→∞} (1+1/x)^{x} is of the indeterminate form 1^{∞}. Before we can use l'Hospital's Rule, we need to bring it into the standard indeterminate form 0/0. For this we use the fact that

lim (ln f(x)) = ln (lim f(x))

So, consider

y = ln (1+1/x)^{x} = x ln (1+1/x) = ln(1+1/x) / (1/x)

Then

lim_{x→∞} y = lim_{x→∞} ln(1+1/x) / (1/x)

which is of the form 0/0, so by l'Hospital's Rule,

lim_{x→∞} y = lim (1/(1+1/x))(-1/x²) / (-1/x²) = lim 1/(1+1/x) = 1.

Therefore,

ln (lim _{x→∞} (1+1/x)^{x}) = 1

lim _{x→∞} (1+1/x)^{x} = e.

Note: this is how you get the continuous compound interest formula!

Blacksburg, VA

Hey David -- we may stairstep mentally ... (1+ 1/1)^1 = 2 ... (1.5)^2 = 2.25 ...

(5/4)^4 = 625/256 ~ 2.5 ... "rule of 72" for doubling is nice ... (11/10)^10 = (1.10)^10

=> 2 at 7th, 2.2, 2.42, 2.68 ... (1.05)^20 => 2 at 14th, 2.1, 2.2, 2.31, 2.43, 2.55,
2.68

pretty nice convergence to 2.7 or "e" ... Best regards, sir :)

Chicago, IL

limit of (1 + 1/x)^{x} as x ---> ∞ is e. **e ≈ 2.71828........**

Middletown, CT

Hi David;

x cannot equal zero because zero cannot be a numerator.

Otherwise, there is no limit.

(negative infinity, 0)U(0, positive infinity)

According to one of my fellow Wyzants, Michael F. of Wilton, Connecticut, "the parentheses ( and ) indicate no endpoints. The symbol ∪ means set union."

I know the answer is e^{1}. Your answer does not make sense to me? I was looking more so on how to get the answer. I'm pretty sure you're giving me the wrong answer.

Are you asking for the limit of the results of this equation?

Or are you asking for the limit of the value of x?

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