Let's solve the limit problems step-by-step. I'll address each question from the image.

Problem 1

$\lim_{x \to 3} (3x^3 + 2x^2 + x + 1)$ Since this is a polynomial, we can directly substitute $x = 3$. $= 3(3)^3 + 2(3)^2 + 3 + 1 = 3 \cdot 27 + 2 \cdot 9 + 3 + 1 = 81 + 18 + 3 + 1 = 103$ Answer for Problem 1: $\lim_{x \to 3} (3x^3 + 2x^2 + x + 1) = 103$.

Problem 2

$\lim_{x \to 3} \frac{x^2 - 5x + 6}{x - 3}$ Factor the numerator: $x^2 - 5x + 6 = (x - 3)(x - 2)$ Thus, we can rewrite the expression as: $\lim_{x \to 3} \frac{(x - 3)(x - 2)}{x - 3}$ Cancel out $x - 3$: $= \lim_{x \to 3} (x - 2) = 3 - 2 = 1$ Answer for Problem 2: $\lim_{x \to 3} \frac{x^2 - 5x + 6}{x - 3} = 1$.

Problem 3

$\lim_{x \to 2} \frac{\sqrt{x^2 + 5} - 3}{x - 2}$ To handle this limit, we use the conjugate: $= \lim_{x \to 2} \frac{\sqrt{x^2 + 5} - 3}{x - 2} \cdot \frac{\sqrt{x^2 + 5} + 3}{\sqrt{x^2 + 5} + 3}$ This gives: $= \lim_{x \to 2} \frac{(x^2 + 5) - 9}{(x - 2)(\sqrt{x^2 + 5} + 3)}$ Simplify the numerator: $= \lim_{x \to 2} \frac{x^2 - 4}{(x - 2)(\sqrt{x^2 + 5} + 3)}$ Factor $x^2 - 4$: $= \lim_{x \to 2} \frac{(x - 2)(x + 2)}{(x - 2)(\sqrt{x^2 + 5} + 3)}$ Cancel $x - 2$: $= \lim_{x \to 2} \frac{x + 2}{\sqrt{x^2 + 5} + 3}$ Substitute $x = 2$: $= \frac{2 + 2}{\sqrt{4 + 5} + 3} = \frac{4}{\sqrt{9} + 3} = \frac{4}{3 + 3} = \frac{4}{6} = \frac{2}{3}$ Answer for Problem 3: $\lim_{x \to 2} \frac{\sqrt{x^2 + 5} - 3}{x - 2} = \frac{2}{3}$.

Problem 4

$\lim_{x \to 5} \frac{4x^2 - 3x - 1}{2x^2 + 2x + 1}$ Since both the numerator and denominator are polynomials of degree 2, we can divide the leading coefficients: $= \frac{4}{2} = 2$ Answer for Problem 4: $\lim_{x \to 5} \frac{4x^2 - 3x - 1}{2x^2 + 2x + 1} = 2$.

Problem 5

$\lim_{x \to 5} \left( \sqrt{x^2 + 2} - \sqrt{x^2 + 1} \right)$ Use the conjugate: $= \lim_{x \to 5} \frac{(\sqrt{x^2 + 2} - \sqrt{x^2 + 1})(\sqrt{x^2 + 2} + \sqrt{x^2 + 1})}{\sqrt{x^2 + 2} + \sqrt{x^2 + 1}}$ This simplifies to: $= \lim_{x \to 5} \frac{(x^2 + 2) - (x^2 + 1)}{\sqrt{x^2 + 2} + \sqrt{x^2 + 1}}$ Simplify the numerator: $= \lim_{x \to 5} \frac{1}{\sqrt{x^2 + 2} + \sqrt{x^2 + 1}}$ Substitute $x = 5$: $= \frac{1}{\sqrt{25 + 2} + \sqrt{25 + 1}} = \frac{1}{\sqrt{27} + \sqrt{26}}$ Answer for Problem 5: $\lim_{x \to 5} \left( \sqrt{x^2 + 2} - \sqrt{x^2 + 1} \right) = \frac{1}{\sqrt{27} + \sqrt{26}}$.

Would you like further details on any specific problem or have any questions?

Related Questions

1. How do we evaluate limits involving square roots without using the conjugate?
2. What is the difference between one-sided and two-sided limits?
3. How do we handle limits that result in indeterminate forms like $\frac{0}{0}$?
4. What are some common strategies for factoring polynomials to simplify limits?
5. How can we evaluate limits at infinity for rational functions?

Tip

When dealing with square roots in limits, using the conjugate is a powerful method to simplify and eliminate square roots.