## Introduction to Limits

The definition of a limit, is denoted as:

$$\lim_{x \to a} f(x) = L$$

states that as ‘x’ gets close to, but not equal to ‘a,’ the function ‘f(x)’ approaches the value ‘L.’ An example for better understanding. If a = 3 and the function f(x) = 4x-2, the limit can be evaluated by direct substitution.

$$\lim_{x \to 3} (4x-2) = 10$$

#### Evaluating Limits for Algebraic Function

1. Product:

$$\lim_{x \to a} f(x) \cdot g(x) = L \cdot K$$

2. Quotient:

$$\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{K}$$

3. Sum or difference:

$$\lim_{x \to a} [f(x) \pm g(x)] = L \pm K$$

4. Scalar multiple:

$$\lim_{x \to a} [bf(x)] = bL$$

5. Power:

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

6. Radical:

$$\lim_{x\to a} \sqrt[n]{f(x)} = \sqrt[n]{l}$$

## Continuity and Discontinuity

A function is said to be continuous at a point ‘a’ if you can directly evaluate the function at ‘a’ without encountering any abrupt changes or jumps. In mathematical terms, if the limit as ‘x’ approaches ‘a’ of the function ‘f(x)’ equals ‘f(a),’ then we conclude that ‘f’ is continuous at ‘a.’ Essentially, there are no gaps or disruptions in the function’s behavior.

$$\lim_{x \to a} f(x) = f(a)$$

Discontinuity arises when the function ‘f’ is not defined at a particular point ‘a.’ This typically occurs when there’s an attempt to divide by zero, resulting in an undefined value at ‘a.’ For instance, if the calculation of ‘f(a)’ leads to a situation where you have a non-zero number divided by zero (non-zero/0), we can assert that the limit as ‘x’ approaches ‘a’ of ‘f(x)’ does not exist.

## Handling Indeterminant Form:

Occasionally, you might encounter situations where ‘f(a)’ equals 0/0, which is termed an “indeterminate form.” It’s essential to understand that an indeterminate form doesn’t imply that the limit cannot be determined; rather, it indicates that simply substituting the value of ‘a’ directly won’t provide a clear answer. In such cases, further algebraic manipulations and simplification are necessary to ascertain the limit.

For example, consider the limit:

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

If you plug in ‘x = 3’ directly, you end up with the indeterminate form 0/0. However, this doesn’t mean the limit doesn’t exist. Instead, it prompts us to explore algebraic techniques, such as factoring the numerator and denominator, to simplify the expression and determine the limit. So in this scenario, we can factor the numerator and denominator.

Let’s factor the expressions separately:

Numerator:

$$(x^2-2x-3) = (x-3)(x+1)$$

Denominator:

$$x^2-9 = (x-3)(x+3)$$

Now we can rewrite the limit and cancel the common factor:

$$\lim_{{x\to3}} \frac{{(x-3)(x+1)}}{{(x-3)(x+3)}} = \lim_{{x\to3}} \frac{{(x+1)}}{{x+3}}$$

Now that we’ve canceled out the common factor, we can evaluate the limit by plugging in ‘x = 3’:

$$\lim_{{x\to3}} \frac{{(x^2-2x-3)}}{{x^2-9}} = \frac{{2}}{3}$$