1. **State the problem:** We need to determine which of the given limits does not yield an indeterminate form. 2. **Recall indeterminate forms:** Common indeterminate forms include $\frac{0}{0}$, $\frac{\infty}{\infty}$, $0 \cdot \infty$, $\infty - \infty$, $0^0$, $1^\infty$, and $\infty^0$. 3. **Analyze each limit:** - A: $$\lim_{x \to 0} \frac{3x^2}{x - \sin x}$$ - As $x \to 0$, numerator $3x^2 \to 0$. - Denominator $x - \sin x \to 0 - 0 = 0$. - So form is $\frac{0}{0}$, which is indeterminate. - B: $$\lim_{x \to 2} \frac{\ln(x)}{x^2 - 5x + 16}$$ - Numerator $\ln(2)$ is a finite nonzero number. - Denominator $2^2 - 5 \cdot 2 + 16 = 4 - 10 + 16 = 10$ (nonzero). - So form is $\frac{\text{finite nonzero}}{\text{nonzero}}$, which is determinate. - C: $$\lim_{x \to \pi} \frac{x - \pi}{\cos x^2}$$ - Numerator $x - \pi \to 0$. - Denominator $\cos(\pi^2)$ is a finite nonzero number (since cosine is continuous and $\cos(\pi^2) \neq 0$). - So form is $\frac{0}{\text{nonzero}} = 0$, determinate. - D: $$\lim_{x \to \infty} \frac{x^x}{x^x}$$ - Both numerator and denominator tend to $\infty$. - Form is $\frac{\infty}{\infty}$, indeterminate. 4. **Conclusion:** Limits B and C are determinate, but only one option is asked. Among the options, B is clearly determinate with no zero denominator or numerator issues. **Final answer:** Option B does not yield an indeterminate form.