One needs some sort of limit for e to start with. A 'standard' limit is
\(\displaystyle \lim_{n->\infty} (1+\frac{1}{n})^n = e\)
Given that, let
n = -\(\displaystyle \frac{1}{2x}\)
then as x goes to zero, n goes to infinity,
1-2x = 1 + \(\displaystyle \frac{1}{n}\),
and
\(\displaystyle \frac{3}{x} = -6 n\).
So we have
\(\displaystyle \lim_{x->0} (1-2x)^{\frac{3}{x}} = \lim_{n->\infty} (1+\frac{1}{n})^{-6n}\)
= \(\displaystyle \lim_{n->\infty} [\frac {1}{(1+\frac{1}{n})}]^{6n}\)
= \(\displaystyle \lim_{n->\infty} [\frac {1}{(1+\frac{1}{n})^n}]^6 = [\frac{1}{e}]^6=e^{-6}\)
I assume you meant that the limit was e-6 and not e6
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