Limit of the Cosine Function
The limit of the cosine of a sequence approaching zero is equal to one. $$ \lim_{a_n \rightarrow 0} \cos a_n = 1 $$
Proof
By hypothesis, the sequence an converges to zero as n approaches infinity.
$$ \lim_{n \rightarrow \infty } a_n = 0 $$
According to the definition of a limit, there exists an index v such that:
$$ -\pi/2 \le a_n \le \pi/2 \quad \text{for all } n > v $$
Note. I chose π/2 because it corresponds to 90°. The cosine of ±90° is equal to zero.
Therefore, for all n > v, the terms of the sequence lie within the interval (-π/2, π/2).
We can now invoke the trigonometric identity relating cosine and sine:
$$ \cos a_n = \pm \sqrt{1 - \sin^2 a_n} $$
Since an stays within (-π/2, π/2), the sine remains strictly less than one in absolute value, and so:
$$ 0 < \sin^2 a_n < 1 $$
Note. The sine at ±90° equals 1.
Thus, the expression \( 1 - \sin^2 a_n \) is strictly positive for all n > v.
It follows that the cosine of an is non-negative:
$$ \cos a_n = \pm \sqrt{1 - \sin^2 a_n} \ge 0 $$
Moreover, since an converges to zero, we have:
$$ \lim_{n \rightarrow \infty} \sin a_n = 0 $$
Hence, \( 1 - \sin^2 a_n \) approaches 1, and therefore:
$$ \lim_{a_n \rightarrow 0} \sqrt{1 - \sin^2 a_n} = \sqrt{1} = 1 $$
Thus, the limit of cos an as an tends to zero is indeed 1.
And this completes the proof.
