We are asked to evaluate the following limits: a) $\lim_{x \to 5} 3x^2 + 2x - 9$ b) $\lim_{x \to 6} \frac{x^2 - 36}{x - 6}$ c) $\lim_{x \to 0} \frac{\sqrt{x^2 + 4} - 2}{x^2}$ d) $\lim_{x \to \infty} \frac{-6x^4 + x^2 + 1}{2x^4 - x}$ e) Examine whether the function $f(x) = \begin{cases} x^3 - 2x + 1 & \text{if } x \le 2 \\ 3x - 2 & \text{if } x > 2 \end{cases}$ is continuous at $x = 2$.

1. Problem Description

We are asked to evaluate the following limits:

a) $\lim_{x \to 5} 3x^2 + 2x - 9$

b) $\lim_{x \to 6} \frac{x^2 - 36}{x - 6}$

c) $\lim_{x \to 0} \frac{\sqrt{x^2 + 4} - 2}{x^2}$

d) $\lim_{x \to \infty} \frac{-6x^4 + x^2 + 1}{2x^4 - x}$

e) Examine whether the function $f(x) = \begin{cases} x^3 - 2x + 1 & \text{if } x \le 2 \\ 3x - 2 & \text{if } x > 2 \end{cases}$ is continuous at $x = 2$.

2. Solution Steps

a) Since the function $3x^2 + 2x - 9$ is a polynomial, we can evaluate the limit by direct substitution:

$\lim_{x \to 5} 3x^2 + 2x - 9 = 3(5^2) + 2(5) - 9 = 3(25) + 10 - 9 = 75 + 10 - 9 = 76$.

b) We can factor the numerator as a difference of squares:

$x^2 - 36 = (x - 6)(x + 6)$.

Then,

$\lim_{x \to 6} \frac{x^2 - 36}{x - 6} = \lim_{x \to 6} \frac{(x - 6)(x + 6)}{x - 6} = \lim_{x \to 6} (x + 6) = 6 + 6 = 12$.

c) We can multiply by the conjugate of the numerator:

$\lim_{x \to 0} \frac{\sqrt{x^2 + 4} - 2}{x^2} = \lim_{x \to 0} \frac{\sqrt{x^2 + 4} - 2}{x^2} \cdot \frac{\sqrt{x^2 + 4} + 2}{\sqrt{x^2 + 4} + 2} = \lim_{x \to 0} \frac{(x^2 + 4) - 4}{x^2(\sqrt{x^2 + 4} + 2)} = \lim_{x \to 0} \frac{x^2}{x^2(\sqrt{x^2 + 4} + 2)} = \lim_{x \to 0} \frac{1}{\sqrt{x^2 + 4} + 2} = \frac{1}{\sqrt{0 + 4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}$.

d) We divide both the numerator and the denominator by the highest power of $x$, which is $x^4$:

$\lim_{x \to \infty} \frac{-6x^4 + x^2 + 1}{2x^4 - x} = \lim_{x \to \infty} \frac{-6 + \frac{1}{x^2} + \frac{1}{x^4}}{2 - \frac{1}{x^3}} = \frac{-6 + 0 + 0}{2 - 0} = \frac{-6}{2} = -3$.

e) For the function to be continuous at $x = 2$, we need $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)$.

$f(2) = (2)^3 - 2(2) + 1 = 8 - 4 + 1 = 5$.

$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^3 - 2x + 1) = (2)^3 - 2(2) + 1 = 8 - 4 + 1 = 5$.

$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x - 2) = 3(2) - 2 = 6 - 2 = 4$.

Since $\lim_{x \to 2^-} f(x) \ne \lim_{x \to 2^+} f(x)$, the function is not continuous at $x = 2$.

3. Final Answer

a) 76

b) 12

c) 1/4

d) -3

e) The function is not continuous at $x = 2$.