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Second example

Probably the best-known examples of finding a limit by squeezing are the proofs of the equalities lim x → 0 sin x x = 1, lim x → 0 1 − cos x x = 0. {\displaystyle {\begin{aligned}&\lim _{x\to 0}{\frac {\sin x}{x}}=1,\\&\lim _{x\to 0}{\frac {1-\cos x}{x}}=0.\end{aligned}}}

The first limit follows by means of the squeeze theorem from the fact that

cos x ≤ sin x x ≤ 1 {\displaystyle \cos x\leq {\frac {\sin x}{x}}\leq 1}

for x close enough to 0. The correctness of which for positive x can be seen by simple geometric reasoning (see drawing) that can be extended to negative x as well. The second limit follows from the squeeze theorem and the fact that

0 ≤ 1 − cos x x ≤ x {\displaystyle 0\leq {\frac {1-\cos x}{x}}\leq x} for x close enough to 0. This can be derived by replacing sin x in the earlier fact by 1 − cos 2 x {\textstyle {\sqrt {1-\cos ^{2}x}}} and squaring the resulting inequality.

These two limits are used in proofs of the fact that the derivative of the sine function is the cosine function. That fact is relied on in other proofs of derivatives of trigonometric functions.