The statement (\lim_{x \to 0} \frac{\sin x}{x} = 1) is correct.

This is a fundamental limit in calculus, essential for deriving the derivative of the sine function ((\frac{d}{dx}\sin x = \cos x)). It can be proven using the squeeze theorem (sandwich theorem):
For small angles (x) (in radians), consider the areas of three shapes in a unit circle:

1. The inscribed triangle: (\frac{1}{2}\sin x)
2. The sector: (\frac{1}{2}x)
3. The circumscribed triangle: (\frac{1}{2}\tan x)

These satisfy (\sin x \leq x \leq \tan x) for (0 < x < \frac{\pi}{2}). Dividing by (\sin x) (positive here) gives (1 \leq \frac{x}{\sin x} \leq \frac{1}{\cos x}). Taking reciprocals (reversing inequalities) gives (\cos x \leq \frac{\sin x}{x} \leq 1). As (x \to 0), (\cos x \to 1), so by the squeeze theorem, (\frac{\sin x}{x} \to 1).

This limit is a cornerstone of trigonometric calculus and is widely used in physics, engineering, and other fields involving periodic functions.

Answer: The statement is correct. (\boxed{1}) (if a numerical answer is expected, though the question is a statement; but since the limit equals 1, this is the value.)

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