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
${\displaystyle \lim_{x\to0}\frac{\sin^{2}\left(\frac{x}{2}\right)}{x^{2}}}$
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
${\displaystyle \lim_{x\to0}\frac{\sin^{2}\left(\frac{x}{2}\right)}{x^{2}}}$
$${\displaystyle \lim_{x\to0}\frac{\sin^{2}\left(\frac{x}{2}\right)}{x^{2}}}$$
$$=\frac 1 4\left(\lim_{y\to 0}\frac{\sin y}y\right)^2$$ where $y=\frac x2$ and $y\to 0$ as $x\to 0$
$$=\frac 1 4$$
$$\lim_{x \to 0}{\sin{\frac{u(x)}{x}}}=1 \tag{1}$$ 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:
$$\lim_{x \to 0 }{\frac{\sin^{2}{\frac{x}{2}}}{x^2}}=\frac{1}{4}. $$