Solution
= Solution
For $0<x<\pi/2$, the usual unit-circle area comparison, with angles measured in radians, gives $\sin x\leq x\leq\tan x$. Hence
$$
\cos x\leq\frac{\sin x}{x}\leq1.
$$
The <cosine> is a <continuous function> at zero, so the <squeeze theorem> yields the right-hand <limit of a function> $1$. The quotient $\sin x/x$ is even, so the left-hand <limit of a function> is the same. Thus \b[the limit is $1$].