Limit of Sine of X over X/Proof 2
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Theorem
- $\displaystyle \lim_{x \mathop \to 0} \frac {\sin x} x = 1$
Proof
From Sine of Zero is Zero:
- $\sin 0 = 0$
From Derivative of Sine Function:
- $D_x \left({\sin x}\right) = \cos x$
Then by Cosine of Zero is One:
- $\cos 0 = 1$
From Derivative of Identity Function:
- $D_x \left({x}\right) = 1$
Thus L'Hôpital's Rule applies and so:
- $\displaystyle \lim_{x \mathop \to 0} \frac {\sin x} x = \lim_{x \mathop \to 0} \frac {D_x \left({\sin x}\right)} {D_x \left({x}\right)} = \lim_{x \mathop \to 0} \frac {\cos x} 1 = \frac 1 1 = 1$
$\blacksquare$