Limit of the Sine Function
The limit of the sine of a sequence an approaching zero is equal to zero. $$ \lim_{a_n \rightarrow 0} \sin a_n = 0 $$
Proof
By assumption, the sequence an converges to zero.
$$ \lim_{n \rightarrow \infty } a_n = 0 $$
Example. Consider the sequence $$ a_n = \frac{1}{n} $$ which converges to zero as n approaches infinity.
By the definition of the limit, there exists an index v such that |an| < ε for all n > v.
From trigonometry, we know that for all positive values of x, the inequality 0 < sin x < x holds. Therefore, we can write:
$$ 0 \le | \sin a_n | = \sin |a_n| \le |a_n| $$
Taking limits as n tends to infinity, we obtain:
$$ \lim_{n \rightarrow \infty } 0 \le \lim_{n \rightarrow \infty } | \sin a_n | = \lim_{n \rightarrow \infty } \sin |a_n| \le \lim_{n \rightarrow \infty } |a_n| $$
Since the sequence an converges to zero, we have:
$$ 0 \le \lim_{n \rightarrow \infty } | \sin a_n | = \lim_{n \rightarrow \infty } \sin |a_n| \le 0 $$
Thus, by the Squeeze Theorem, the limit of sin an must also be zero:
$$ \lim_{n \rightarrow \infty } | \sin a_n | = \lim_{n \rightarrow \infty } \sin |a_n| = 0 $$
Therefore, if the sequence an approaches zero as n→∞, then the limit of sin(an) is also zero.
$$ \lim_{a_n \rightarrow 0} \sin a_n = \sin 0 = 0 $$
And that completes the proof.
