1. **Problem a:** Find the limit $\lim_{x \to 1} x^2 - 3x + 5$. 2. **Formula and rules:** For polynomial functions, limits at a point can be found by direct substitution because polynomials are continuous everywhere. 3. **Calculate:** Substitute $x = 1$ into the expression: $$1^2 - 3 \times 1 + 5 = 1 - 3 + 5$$ 4. **Simplify:** $$1 - 3 + 5 = (1 - 3) + 5 = -2 + 5 = 3$$ 5. **Answer for a:** $$\lim_{x \to 1} x^2 - 3x + 5 = 3$$ 6. **Problem b:** Find the limit $\lim_{u \to \infty} \frac{3u^3 + 14u^2 + 11u}{u^3 - 1}$. 7. **Formula and rules:** For rational functions where numerator and denominator are polynomials, the limit at infinity depends on the degrees of the polynomials. - If degrees are equal, limit is ratio of leading coefficients. - If numerator degree > denominator degree, limit is $\pm \infty$. - If numerator degree < denominator degree, limit is 0. 8. **Calculate:** Both numerator and denominator have degree 3. Leading coefficients are 3 (numerator) and 1 (denominator). 9. **Simplify by dividing numerator and denominator by $u^3$:** $$\frac{3u^3 + 14u^2 + 11u}{u^3 - 1} = \frac{\cancel{u^3}(3 + \frac{14}{u} + \frac{11}{u^2})}{\cancel{u^3}(1 - \frac{1}{u^3})}$$ 10. **Take the limit as $u \to \infty$:** Terms with $\frac{1}{u}$ and higher powers go to 0. $$\lim_{u \to \infty} \frac{3 + 0 + 0}{1 - 0} = 3$$ 11. **Answer for b:** $$\lim_{u \to \infty} \frac{3u^3 + 14u^2 + 11u}{u^3 - 1} = 3$$