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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

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

2 Answers 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, http://www.ies.co.jp/math/java/calc/LimSinX/LimSinX.html2012-11-20
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$\lim_{x \to 0}{\sin{\frac{u(x)}{x}}}=1 \tag{1}$ 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}. $