# finding a function limit

I was wondering how to find the limit of the following function, without using the l'hopital rule or any advanced theorams. no need for epsilon-delta defintion

${\displaystyle \lim_{x\to0}\frac{\sin^{2}\left(\frac{x}{2}\right)}{x^{2}}}$

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$${\displaystyle \lim_{x\to0}\frac{\sin^{2}\left(\frac{x}{2}\right)}{x^{2}}}$$

$$=\frac 1 4\left(\lim_{y\to 0}\frac{\sin y}y\right)^2$$ where $y=\frac x2$ and $y\to 0$ as $x\to 0$

$$=\frac 1 4$$

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@sonyjimbo, $m\cdot 0=0$ for any finite $m,$ so $m$ is undetermined and $y\to 0$ means $y\ne 0$ –  lab bhattacharjee Nov 20 '12 at 16:42
–  lab bhattacharjee Nov 20 '12 at 16:45

$$\lim_{x \to 0}{\sin{\frac{u(x)}{x}}}=1 \tag{1}$$ when $\displaystyle \lim_{x \to 0}{u(x)}=0$.

$$\lim_{x \to 0}{\frac{\left(\sin{\frac{x}{2}}\right)^{2}}{x^2}}=\lim_{x \to 0}{\left(\frac{\sin{\frac{x}{2}}}{x}\right)^{2}}=\lim_{x \to 0}{\left(\frac{\sin{\frac{x}{2}}}{\frac{x}{2}}\right)^{2}} \cdot \frac{1}{4}$$ and using $(1)$ we obtain that:

$$\lim_{x \to 0 }{\frac{\sin^{2}{\frac{x}{2}}}{x^2}}=\frac{1}{4}.$$

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