lim ((sin^2)x)/x
x->0
This is a tricky question... Can anybody lend a helping hand?
$\displaystyle \lim_{x \to 0} \frac{\frac{1}{2+x} - \frac{1}{2}}{x}$
Find common denominator in the numerator
$\displaystyle \lim_{x \to 0} \frac{\frac{2-2-x}{2(2+x)}}{x} = \frac{\frac{-x}{4+2x}}{x} = \frac{-x}{x(4+2x)}$
which equals
$\displaystyle \lim_{x \to 0}~\frac{-1}{4+2x}~=~\frac{-1}{4}$