Let's work through the problems shown in the image:

1. $\lim_{x \to 5} \left( 4x^2 + 2x - 1 \right) = \ldots$

This is a limit where direct substitution can be applied, as the function is continuous at $x = 5$. So, substitute $x = 5$ directly:

$\lim_{x \to 5} \left( 4x^2 + 2x - 1 \right) = 4(5)^2 + 2(5) - 1 = 4(25) + 10 - 1 = 100 + 10 - 1 = 109$ Thus, the answer is B. 109.

2. $\lim_{x \to 4} \frac{x^2 - 2x - 8}{x - 4} = \ldots$

Here, we have a rational function. First, factor the numerator:

$x^2 - 2x - 8 = (x - 4)(x + 2)$

Now, cancel out the common factor of $(x - 4)$:

$\frac{(x - 4)(x + 2)}{x - 4} = x + 2$

Now, apply the limit by substituting $x = 4$:

$\lim_{x \to 4} (x + 2) = 4 + 2 = 6$ Thus, the answer is E. 6.

3. $\lim_{x \to \infty} \frac{3x^3 - 2x - 10}{4x^3 - 2x^2 - 5x^3} = \ldots$

We deal with infinity limits by looking at the highest powers of $x$ in both the numerator and denominator. The highest power in the numerator is $3x^3$, and in the denominator, it's $-5x^3$. So, we simplify the limit as follows:

$\lim_{x \to \infty} \frac{3x^3 - 2x - 10}{4x^3 - 2x^2 - 5x^3} = \lim_{x \to \infty} \frac{3x^3}{-5x^3} = \frac{3}{-5} = -\frac{3}{5}$ Thus, the answer is B. -$\frac{3}{5}$.

4. Find the asymptote of the function $f(x) = \frac{x + 3}{x - 5}$

To find the vertical asymptote, set the denominator equal to zero:

$x - 5 = 0 \implies x = 5$

For the horizontal asymptote, examine the degree of the numerator and the denominator. Both are degree 1 (linear), so the horizontal asymptote is given by the ratio of the leading coefficients:

$\text{Horizontal asymptote: } \frac{1}{1} = 1$ Thus, the asymptote is $y = 1$.

5. Check the continuity of the function ( f(x) = \begin{cases}

x & \text{if } x < 0 \ x^2 & \text{if } 0 \leq x \leq 1 \ 2 - x & \text{if } x > 1 \end{cases} ) To check for continuity at $x = 0$ and $x = 1$, we must check that the left-hand limit, right-hand limit, and the value of the function match.

• At $x = 0$:

  • Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x = 0$
  • Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^2 = 0^2 = 0$
  • Function value: $f(0) = 0^2 = 0$

  Thus, $f(x)$ is continuous at $x = 0$.

• At $x = 1$:

  • Left-hand limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^2 = 1^2 = 1$
  • Right-hand limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2 - x) = 2 - 1 = 1$
  • Function value: $f(1) = 1^2 = 1$

  Thus, $f(x)$ is continuous at $x = 1$.

So, the function is continuous on its entire domain.

Would you like further details or have any questions?

Follow-up questions:

1. What is the significance of the degree of polynomials in determining limits?
2. How do you solve limits involving infinity?
3. What are the steps to finding vertical and horizontal asymptotes?
4. Can a function have both a vertical and a horizontal asymptote?
5. How do you check for differentiability along with continuity?

Tip: When simplifying limits, always check for indeterminate forms like $\frac{0}{0}$ to decide on factoring or L'Hopital's Rule.