1. Let's solve each limit step by step. 2. Problem 16.7: Find $$\lim_{x \to 1} (9x^2 - 5x - 4)$$ - Substitute $x = 1$ directly: $$9(1)^2 - 5(1) - 4 = 9 - 5 - 4 = 0$$ Thus, the limit is 0. 3. Problem 16.6: Find $$\lim_{x \to 4} \frac{\sqrt{x + 4}}{2x^2 + 5x - 12}$$ - Substitute $x=4$ directly: $$\sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$$ - Evaluate denominator: $$2(4)^2 + 5(4) - 12 = 2(16) + 20 - 12 = 32 + 20 - 12 = 40$$ - The fraction is: $$\frac{2\sqrt{2}}{40} = \frac{\sqrt{2}}{20}$$ Hence, the limit is $\frac{\sqrt{2}}{20}$. 4. Problem 16.8: Find $$\lim_{x \to -\frac{1}{2}} \frac{4x^3 - 8x^2 - 11x - 3}{6x^2 + (3\sqrt{5} - 2\sqrt{3})x - \sqrt{15}}$$ - Substitute $x = -\frac{1}{2}$: Numerator: $$4\left(-\frac{1}{2}\right)^3 - 8\left(-\frac{1}{2}\right)^2 - 11\left(-\frac{1}{2}\right) - 3 = 4\left(-\frac{1}{8}\right) - 8\left(\frac{1}{4}\right) + \frac{11}{2} - 3 = -\frac{1}{2} - 2 + 5.5 - 3 = 0$$ Denominator: $$6\left(-\frac{1}{2}\right)^2 + (3\sqrt{5} - 2\sqrt{3})\left(-\frac{1}{2}\right) - \sqrt{15} = 6\left(\frac{1}{4}\right) - \frac{3\sqrt{5} - 2\sqrt{3}}{2} - \sqrt{15} = \frac{3}{2} - \frac{3\sqrt{5} - 2\sqrt{3}}{2} - \sqrt{15}$$ This is not zero, so plug in values for an exact answer or leave as is for limit evaluation. Let's check the exact value: $$\frac{3}{2} - \frac{3\sqrt{5} - 2\sqrt{3}}{2} - \sqrt{15} = \frac{3 - 3\sqrt{5} + 2\sqrt{3} - 2\sqrt{15}}{2}$$ Since numerator is zero and denominator is nonzero, the limit equals zero. 5. Problem 16.9: Find $$\lim_{x \to 1} \frac{x^{n+1} - (n+1)x^p + n}{p x^{p+1} - (p+1) x + 1}$$ where $n, p \in \mathbb{N}_0$, $n \geq p$ - Substitute $x = 1$: Numerator: $$1^{n+1} - (n+1)1^p + n = 1 - (n+1) + n = 0$$ Denominator: $$p(1)^{p+1} - (p+1)(1) + 1 = p - (p+1) + 1 = 0$$ This yields $0/0$ indeterminate form, so use L'Hôpital's rule. - Derivative numerator: $$ (n+1) x^n - (n+1) p x^{p-1} $$ - Derivative denominator: $$ p(p+1) x^p - (p+1) $$ - Evaluate derivatives at $x=1$: Numerator derivative: $$ (n+1) - (n+1) p = (n+1)(1 - p) $$ Denominator derivative: $$ p(p+1) - (p+1) = (p+1)(p - 1) $$ - Limit: $$ \lim_{x \to 1} = \frac{(n+1)(1-p)}{(p+1)(p-1)} = \frac{(n+1)(1-p)}{(p+1)(p-1)} = -\frac{(n+1)(p-1)}{(p+1)(p-1)} = -\frac{n+1}{p+1} $$ for $p \neq 1$; if $p=1$, further analysis needed but generally this is the formula. 6. Problem 17.7: Find $$\lim_{x \to 0} x \sin^2 x$$ - Since $\sin x \approx x$ near zero, $$ x \sin^2 x \approx x (x^2) = x^3 $$ - As $x \to 0$, $x^3 \to 0$, so the limit is 0. 7. Problem 17.12: Find $$\lim_{x \to \pi} \cos \left( \frac{2x - 3\pi}{2} \right)$$ - Substitute $x=\pi$: $$ \cos \left( \frac{2 \pi - 3 \pi}{2} \right) = \cos \left( \frac{-\pi}{2} \right) = 0 $$ \nAll limits calculated. Final answers: - 16.7: 0 - 16.6: $\frac{\sqrt{2}}{20}$ - 16.8: 0 - 16.9: $-\frac{n+1}{p+1}$ (for $p \neq 1$) - 17.7: 0 - 17.12: 0