They're kinda the same thing...
\frac{1}{0} = \infty
\sin^2(0) = 0
so
\frac{\sin^2(0)}{0} = \underbrace{\frac{0}{0}}_\text{ind.} = \frac{1}{0}\cdot\frac{0}{1} = \underbrace{\infty\cdot0}_\text{ind.} = \frac{1}{0}\cdot\frac{\sin^2(0)}{1}
Use l'Hôpital's rule to find the value for this limit.
If \lim\limits_{x\to\alpha}\frac{f(x)}{g(x)} = \frac{0}{0} \mathrm{or} \pm\frac{\infty}{\infty}, then \lim\limits_{x\to\alpha}\frac{f(x)}{g(x)} = \lim\limits_{x\to\alpha}\frac{f'(x)}{g'(x)}
so, define functions...
f: x \ {\mapsto}\ \sin\left(\frac{1}{2} \, x\right)^{2}
g: x \ {\mapsto}\ x^{2}
Differentiate until you aren't going to get an indeterminate value (that is, make sure that the denominator does not equal zero).
\frac{\mathrm{d}g}{\mathrm{d}x} = 2 \, x
\frac{\mathrm{d}^{2}g}{\mathrm{d}x^{2}} = 2
So now the denominator will never equal zero. Differentiate f the same number of times.
\frac{\mathrm{d}f}{\mathrm{d}x} = \sin\left(\frac{1}{2} \, x\right) \cos\left(\frac{1}{2}<br />
\, x\right)
\frac{\mathrm{d}^{2}f}{\mathrm{d}x^{2}} = -\frac{1}{2} \, \sin\left(\frac{1}{2} \, x\right)^{2} +<br />
\frac{1}{2} \, \cos\left(\frac{1}{2} \, x\right)^{2}
Now divide the functions
\frac{f''(x)}{g''(x)} = -\frac{1}{4} \, \sin\left(\frac{1}{2} \, x\right)^{2} +<br />
\frac{1}{4} \, \cos\left(\frac{1}{2} \, x\right)^{2}
simplifies to
\frac{f''(x)}{g''(x)} = \frac{1}{2} \, \cos\left(\frac{1}{2} \, x\right)^{2} -<br />
\frac{1}{4}
now take the limit as x tends to zero
\lim\limits_{x\to0}\frac{1}{2} \, \cos\left(\frac{1}{2} \, x\right)^{2} -<br />
\frac{1}{4} = \frac{1}{2} \, \cos\left(0\right)^{2} -<br />
\frac{1}{4} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}
And, to bring it all together:
\lim\limits_{x\to0}\dfrac{\sin^{2}\left(\frac{x}{2}\right)}{x^{2}} = \lim\limits_{x\to0}\frac{1}{2} \, \cos\left(\frac{1}{2} \, x\right)^{2} -<br />
\frac{1}{4} = \frac{1}{4}
I hope that helps!