Lim x-->∞ (1+^{1}/_{x})^{x}
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!
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