Question

Evaluate the limit $$ \lim_{x \to \infty} \frac{\sin x}{x} $$

• A. 1
• B. -1
• C. \( \infty \)
• D. zero

Hint:

When the numerator of a fraction is bounded (like \( \sin x \)) and the denominator grows without bound (like \( x \)), the limit of the fraction approaches zero.

Answer 1

The Correct Option is D

We are asked to evaluate the limit: \[ \lim_{x \to \infty} \frac{\sin x}{x} \] As \( x \) approaches infinity, we know that \( \sin x \) oscillates between -1 and
(1) However, \( x \) increases without bound. Hence, the fraction \( \frac{\sin x}{x} \) will approach zero because the numerator remains bounded while the denominator grows infinitely large.
More formally, using the fact that:
\[ -1 \leq \sin x \leq 1 \] we can write:
\[ \frac{-1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} \] As \( x \to \infty \), both \( \frac{-1}{x} \) and \( \frac{1}{x} \) approach 0, by the Squeeze Theorem:
\[ \lim_{x \to \infty} \frac{\sin x}{x} = 0 \]
Conclusion: The value of the limit is zero.