Concept of Limit Function

Understanding Limits Intuitively

Imagine you are walking towards your house door. The closer you get to the door, the clearer you can see the details of the door. In mathematics, limits work in a similar way. Limits describe the value that a function approaches when its input variable approaches a certain value.

The concept of limits is very fundamental in calculus because it becomes the foundation for understanding derivatives, integrals, and function continuity. Limits help us understand function behavior around certain points, even when the function is not defined exactly at that point.

Approach Through Value Tables

To understand limits more concretely, let's see how function values change when the variable approaches a certain point. Suppose we have function $f(x) = x + 2$ and want to see what happens when $x$ approaches 3.

From the table above, we can see that when $x$ approaches 3 from the left (values $x < 3$) and from the right (values $x > 3$), the value of $f(x)$ approaches 5. This approaching value is called the limit.

Formal Definition of Limits

Mathematically, limits can be defined as follows:

This definition is read as "the limit of $f(x)$ as $x$ approaches $c$ equals $L$".

The conditions for this limit to exist are:

• Left limit and right limit must exist
• Left limit must be equal to the right limit
• The limit value is $L$

More formally, left and right limits can be written as:

If both limits are equal, then $\lim_{x \to c} f(x) = L$. If they are different, then the limit does not exist.

Application of Limits

Simple Example

Find $\lim_{x \to 4} (2x - 1)$.

Solution:

Since the function $f(x) = 2x - 1$ is continuous at $x = 4$, we can directly substitute:

Example with Indeterminate Form

Find $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$.

Solution:

If we substitute $x = 2$ directly, we get the indeterminate form $\frac{0}{0}$. We need to simplify first by factoring:

Since we are calculating the limit when $x$ approaches 2 (not equals 2), then $x \neq 2$ and we can cancel $(x - 2)$:

Basic Properties of Limits

Some important properties that facilitate limit calculations:

1. Linearity Property:

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

2. Multiplication Property:

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

3. Division Property:

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

Exercises

1. Find $\lim_{x \to 3} (x^2 + 2x - 1)$

2. Find $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$

3. Find $\lim_{x \to 0} \frac{\sin x}{x}$ (use trigonometric limit theorem)

4. If $f(x) = \begin{cases} x + 1, & x < 2 \\ 3x - 2, & x \geq 2 \end{cases}$, find $\lim_{x \to 2} f(x)$

Answer Key

1. Solution:

   Since polynomial functions are continuous at all points, we can substitute directly:

   $$\lim_{x \to 3} (x^2 + 2x - 1) = 3^2 + 2(3) - 1 = 9 + 6 - 1 = 14$$

2. Solution:

   Direct substitution yields the form $\frac{0}{0}$. We factor first:

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

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

   Since $x$ approaches 1 (not equal to 1), then $x \neq 1$ and we can cancel $(x - 1)$:

   $$= \lim_{x \to 1} (x + 1) = 1 + 1 = 2$$

3. Solution:

   This is a fundamental trigonometric limit that is very important in calculus. This limit cannot be solved by direct substitution because it would yield the form $\frac{0}{0}$. However, based on the proven trigonometric limit theorem:

   $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$

   Note: $x$ is in radians, not degrees.

4. Solution:

   For piecewise functions (defined with different rules), we must check left and right limits separately:

   Left limit (when $x$ approaches 2 from the left, so $x < 2$):

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

   Right limit (when $x$ approaches 2 from the right, so $x \geq 2$):

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

   Since $\lim_{x \to 2^-} f(x) = 3 \neq 4 = \lim_{x \to 2^+} f(x)$, then $\lim_{x \to 2} f(x)$ does not exist.