Squeeze theorem applied to sin(x)/x

First evaluate the limits of the two bounding functions as $x \to 0$. Since $1 - \frac{x^2}{2}$ is continuous at $x=0$, direct substitution gives $\lim_{x\to 0}\left(1 - \frac{x^2}{2}\right) = 1 - 0 = 1$. The upper bound is the constant function $1$, so trivially $\lim_{x\to 0} 1 = 1$.

Since $1 - \frac{x^2}{2} \le \frac{\sin(x)}{x} \le 1$ near $x=0$ (excluding $x=0$ itself), and both the lower bound $1 - \frac{x^2}{2}$ and upper bound $1$ approach the same value $1$ as $x \to 0$, the Squeeze Theorem guarantees that the function trapped between them, $\frac{\sin(x)}{x}$, must also approach $1$. Therefore $\lim_{x\to 0} \frac{\sin(x)}{x} = 1$.