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

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

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

$\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\lim_{x \to c} f(x)}{\lim_{x \to c} g(x)}$, provided $\lim_{x \to c} g(x) \neq 0$

If $\lim_{x \to c} f(x) = 0$ and $\lim_{x \to c} g(x) = 0$, then $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

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

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

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

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

If $f(x) \leq g(x) \leq h(x)$ for all x near c, and $\lim_{x \to c} f(x) = L = \lim_{x \to c} h(x)$, then $\lim_{x \to c} g(x) = L$.

If $f(x) \le g(x) \le h(x)$ for all x near a, and $\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L$, then $\lim_{x \to a} g(x) = L$.

If f is continuous on [a, b], then for any value N between f(a) and f(b), there exists a c in (a, b) such that f(c) = N.

If f is continuous on a closed interval [a, b], then f has both a maximum and a minimum value on that interval.

If $\lim_{x \to c} f(x) = 0$ and $\lim_{x \to c} g(x) = 0$ (or both approach infinity), and if $f'(x)$ and $g'(x)$ exist, then $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$.

If you can bound a function between two other functions that have the same limit, then the function in the middle must also have the same limit.

To show that a continuous function takes on a specific value within a given interval.

To guarantee the existence of maximum and minimum values for a continuous function on a closed interval.

When evaluating limits that result in indeterminate forms such as 0/0 or ∞/∞.

The limit of a constant times a function is the constant times the limit of the function: $\lim_{x \to c} [k \cdot f(x)] = k \cdot \lim_{x \to c} f(x)$

The limit of a sum (or difference) is the sum (or difference) of the limits: $\lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

1. Locate 'a' on the x-axis. 2. Follow the graph as x approaches 'a' from both sides. 3. Identify the y-value that the graph approaches. 4. That y-value is the limit.

1. Look at x-values approaching 'a' from both sides. 2. Observe the corresponding y-values. 3. If y-values approach the same number, that's the limit.

1. Try to factor and simplify the expression. 2. If simplification doesn't work, consider L'Hôpital's Rule. 3. Take the derivative of the numerator and denominator. 4. Evaluate the limit again.

1. Multiply the numerator and denominator by the conjugate of the expression containing the radical. 2. Simplify the expression. 3. Evaluate the limit.

1. Divide both numerator and denominator by the highest power of x in the denominator. 2. Evaluate the limit as x approaches infinity. Terms like 1/x approach 0.

1. Find the limit from the left. 2. Find the limit from the right. 3. If the left-hand limit equals the right-hand limit, the limit exists and is equal to that value.

1. Find two functions, f(x) and h(x), such that f(x) <= g(x) <= h(x). 2. Find the limits of f(x) and h(x) as x approaches a. 3. If both limits are equal to L, then the limit of g(x) as x approaches a is also L.

1. Find $\lim_{x \to a} g(x) = L$. 2. Find $\lim_{x \to L} f(x)$. 3. If this limit exists, it is the limit of the composite function.

1. Determine which piece of the function applies as x approaches the target value. 2. Evaluate the limit using that piece of the function. 3. If the target value is the boundary, check both left and right limits.

1. First, try direct substitution. 2. If direct substitution results in an indeterminate form, then use algebraic manipulation techniques.