Originally Posted by
TheEmptySet Another way to think about this is in terms of the derivative.
Let
$\displaystyle f(x)=\frac{\sin(x)}{x}=\frac{\sin(x)-0}{x-0}$
so the limits is
$\displaystyle \lim_{x \to 0}[f(x)]^2=\left( \lim_{x \to 0} f(x)\right)\left( \lim_{x \to 0} f(x)\right)$
but each factor is just the definition of the derivative evaluates at zero
but
$\displaystyle \frac{d}{dx}\sin(x) \bigg|_{x=0}=1$
$\displaystyle \lim_{x \to 0}[f(x)]^2=\left( \lim_{x \to 0} f(x)\right)\left( \lim_{x \to 0} f(x)\right)=1\cdot 1=1$