1. Direct substitution: $(2)^2 + 3(2) - 5 = 4 + 6 - 5 = 5$. Therefore, the limit is 5.

1. Direct substitution yields 0/0. 2. Factor the numerator: $x^2 - 9 = (x - 3)(x + 3)$. 3. Simplify: $\frac{(x - 3)(x + 3)}{x - 3} = x + 3$. 4. Evaluate the limit: $lim_{x o 3} (x + 3) = 3 + 3 = 6$.

1. Direct substitution yields 0/0. 2. Rationalize the numerator by multiplying by the conjugate: $\frac{\sqrt{x + 4} - 2}{x} cdot \frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2}$. 3. Simplify: $\frac{x + 4 - 4}{x(\sqrt{x + 4} + 2)} = \frac{x}{x(\sqrt{x + 4} + 2)} = \frac{1}{\sqrt{x + 4} + 2}$. 4. Evaluate the limit: $lim_{x o 0} \frac{1}{\sqrt{x + 4} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{4}$.

1. Direct substitution: 2(5)^2 - 3 = 2(25) - 3 = 50 - 3 = 47. Therefore, the limit is 47.

1. Direct substitution yields 0/0. 2. Factor the numerator: $x^3 + 8 = (x + 2)(x^2 - 2x + 4)$. 3. Simplify: $\frac{(x + 2)(x^2 - 2x + 4)}{x + 2} = x^2 - 2x + 4$. 4. Evaluate the limit: $lim_{x o -2} (x^2 - 2x + 4) = (-2)^2 - 2(-2) + 4 = 4 + 4 + 4 = 12$.

1. Direct substitution yields 0/0. 2. Rationalize the numerator: $\frac{\sqrt{x} - 2}{x - 4} cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2}$. 3. Simplify: $\frac{x - 4}{(x - 4)(\sqrt{x} + 2)} = \frac{1}{\sqrt{x} + 2}$. 4. Evaluate the limit: $lim_{x o 4} \frac{1}{\sqrt{x} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{4}$.

1. Direct substitution yields 0/0. 2. Factor the numerator: $x^2 + 4x - 5 = (x - 1)(x + 5)$. 3. Simplify: $\frac{(x - 1)(x + 5)}{x - 1} = x + 5$. 4. Evaluate the limit: $lim_{x o 1} (x + 5) = 1 + 5 = 6$.

1. Direct substitution yields 0/0. 2. Expand: $\frac{x^2+10x+25 - 25}{x} = \frac{x^2+10x}{x}$. 3. Simplify: $\frac{x(x+10)}{x} = x+10$. 4. Evaluate the limit: $lim_{x \to 0} (x+10) = 0+10 = 10$.

1. Direct substitution: 5(2) - 3(2)^3 = 10 - 3(8) = 10 - 24 = -14. Therefore, the limit is -14.

1. Direct substitution yields 0/0. 2. Factor the numerator: $x^2 - 1 = (x - 1)(x + 1)$. 3. Simplify: $\frac{(x - 1)(x + 1)}{x + 1} = x - 1$. 4. Evaluate the limit: $lim_{x o -1} (x - 1) = -1 - 1 = -2$.

$lim_{x o c} [f(x) + g(x)] = lim_{x o c} f(x) + lim_{x o c} g(x)$

$lim_{x o c} [f(x) - g(x)] = lim_{x o c} f(x) - lim_{x o c} g(x)$

$lim_{x o c} [k cdot f(x)] = k cdot lim_{x o c} f(x)$

$lim_{x o c} [f(x) cdot g(x)] = lim_{x o c} f(x) cdot lim_{x o c} g(x)$

$lim_{x o c} \frac{f(x)}{g(x)} = \frac{lim_{x o c} f(x)}{lim_{x o c} g(x)}$, if lim_{x o c} g(x) eq 0

$lim_{x o c} [f(x)]^n = [lim_{x o c} f(x)]^n$

$lim_{x o c} \sqrt[n]{f(x)} = \sqrt[n]{lim_{x o c} f(x)}$

$lim_{x \to c} k = k$

$lim_{x \to c} x = c$

$lim_{x \to c} x^n = c^n$

First attempt to evaluate a limit by plugging in the value that x approaches. If it results in a defined value, that is the limit.

Use algebraic manipulation (factoring, rationalizing, etc.) or other techniques to simplify the expression before evaluating the limit.

Limit laws allow you to break down complex limits into simpler parts by dealing with sums, differences, products, quotients, and powers separately.

The Quotient Rule can only be applied if the limit of the denominator is not zero. Otherwise, the limit does not exist or requires further analysis.

If the function is a constant, the limit is simply that constant value, regardless of what x approaches.

One-sided limits examine the behavior of a function as x approaches a value from the left (x -> c-) or from the right (x -> c+).

For a two-sided limit to exist at a point, both the left-hand limit and the right-hand limit must exist and be equal at that point.

Attempt direct substitution. If it yields a defined value, the limit is found. If it yields an indeterminate form, further techniques are required.

Limit laws can be applied when the individual limits of the functions involved exist. They help simplify complex expressions into manageable components.

For a function to be continuous at a point, the limit must exist at that point, the function must be defined at that point, and the limit must equal the function's value at that point.