Question

find the limit. (in the limit, direct substitution applies?
\\(\lim_{x \to \infty} \left( \frac{\sin^2(x)}{x^2 + 8} \
ight)\\))

Explanation:

Step1: Bound the numerator

We know that for all real $x$, $0 \leq \sin^2(x) \leq 1$.

Step2: Create inequality for the fraction

Divide all parts of the inequality by $x^2 + 8$ (which is always positive for real $x$):
$$0 \leq \frac{\sin^2(x)}{x^2 + 8} \leq \frac{1}{x^2 + 8}$$

Step3: Evaluate limits of bounds

Compute $\lim_{x \to \infty} 0 = 0$, and $\lim_{x \to \infty} \frac{1}{x^2 + 8} = 0$ (since the denominator grows without bound).

Step4: Apply Squeeze Theorem

Since the function is squeezed between two functions both approaching 0, its limit is also 0.

Answer:

$0$