The image shows a set of calculus exercises involving limits of functions. Let’s solve the problems one by one:

Problem 1:

a) $\lim_{x \to -2} \left( x^2 + 2x - 1 \right)$

Substitute $x = -2$ directly (since this is a polynomial, it is continuous):

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

Answer: $-1$

b) $\lim_{t \to -1} \frac{\sqrt{1 - 2t}}{(3t + 2)^3}$

1. First, substitute $t = -1$ into the numerator and denominator:

   • Numerator: $\sqrt{1 - 2(-1)} = \sqrt{1 + 2} = \sqrt{3}$,
   • Denominator: $(3(-1) + 2)^3 = (-3 + 2)^3 = (-1)^3 = -1$.

2. Result: $\lim_{t \to -1} \frac{\sqrt{1 - 2t}}{(3t + 2)^3} = \frac{\sqrt{3}}{-1} = -\sqrt{3}.$

Answer: $-\sqrt{3}$

c) $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$

1. Factorize the numerator: $x^2 - 9 = (x - 3)(x + 3).$

2. Simplify the expression: $\frac{x^2 - 9}{x - 3} = x + 3 \quad \text{for } x \neq 3.$

3. Substitute $x = 3$: $\lim_{x \to 3} (x + 3) = 3 + 3 = 6.$

Answer: $6$

d) $\lim_{x \to 3} \frac{x^4 - 18x^2 + 81}{(x - 3)^2}$

1. Factorize $x^4 - 18x^2 + 81$:

   • Let $y = x^2$, so the expression becomes $y^2 - 18y + 81 = (y - 9)^2$,
   • Substituting back: $x^4 - 18x^2 + 81 = (x^2 - 9)^2 = [(x - 3)(x + 3)]^2.$

2. Simplify: $\frac{(x^4 - 18x^2 + 81)}{(x - 3)^2} = \frac{[(x - 3)(x + 3)]^2}{(x - 3)^2} = (x + 3)^2.$

3. Substitute $x = 3$: $\lim_{x \to 3} (x + 3)^2 = (3 + 3)^2 = 6^2 = 36.$

Answer: $36$

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

1. Simplify the expression by dividing each term by $x^2$ (for $x \neq 0$): $\frac{x^4 + 2x^2 - x}{x^2} = x^2 + 2 - \frac{1}{x}.$

2. As $x \to 0$, the term $-\frac{1}{x}$ diverges to $-\infty$. Therefore, the limit does not exist.

Answer: Does not exist (diverges to $-\infty$)

f) $\lim_{t \to 7} \frac{\sqrt{(t - 7)^3}}{(t - 7)}$

1. Simplify the square root: $\sqrt{(t - 7)^3} = (t - 7)^{3/2}.$

2. Simplify the fraction: $\frac{\sqrt{(t - 7)^3}}{(t - 7)} = (t - 7)^{3/2 - 1} = (t - 7)^{1/2}.$

3. As $t \to 7$, $t - 7 \to 0$, so: $\lim_{t \to 7} (t - 7)^{1/2} = 0.$

Answer: $0$

Summary of Results:

a) $-1$
b) $-\sqrt{3}$
c) $6$
d) $36$
e) Does not exist
f) $0$

Do you want a detailed explanation of any of the solutions?

Related Questions:

1. How do you simplify limits involving square roots?
2. What are the rules for factoring polynomial expressions to solve limits?
3. What is L'Hopital's Rule, and when should it be applied?
4. How do you determine if a limit diverges or converges?
5. Why do some limits not exist?

Tip: For rational functions, always check for removable discontinuities by factoring the numerator and denominator!