Finding the limit of a function is a fundamental concept in calculus that helps us understand the behavior of functions as they approach specific values. Limits describe how a function behaves near a point, even if the function is not defined at that exact point.

This concept is crucial for understanding continuity, derivatives, and integrals. Mastering limit techniques allows students to analyze function behavior, solve optimization problems, and explore rates of change in various fields such as physics, engineering, and economics.

Table of Content

- What are Limits?
- Types of Limits
- Importance in Calculus
- Limit of a Function
- Left-Hand Limits and Right-Hand Limits
- Finding Limit using Table
- Steps to Find The Limit of a Function
- Solved Examples
- Practice Questions

## What is Limit?

Limits are a fundamental concept in calculus that describes how a function behaves as its input approaches a particular value. They are crucial for understanding and defining derivatives, integrals, and continuity.

For a function f(x) and a point aaa, the limit of f(x) as x approaches aaa is a value L if f(x) gets arbitrarily close to L as x approaches a from both sides. This is written as lim_{x→a}f(x)=L.

**Types of Limits**

**Types of Limits**

When f(x) approaches a finite value L.**Finite Limits:**When f(x) grows without bound as x approaches aaa.**Infinite Limits:**When x approaches infinity or negative infinity, and f(x) approaches a finite value or infinity.**Limits at Infinity:**

### Limit of a Function

Limits are fundamental to calculus, forming the basis for the concept of derivatives.

- Consider a function f(x) and a constant ‘a’:
- We examine values of x that are close to, but not equal to, ‘a’.
- We observe the corresponding values of f(x) as x approaches ‘a’.
- If f(x) consistently gets closer to a specific value A as x gets closer to ‘a’, we say that A is the limit of f(x) as x approaches ‘a’.

**Finding Limit using Table**

**Finding Limit using Table**

Using a table to find the limit of a function involves approximating the value of the function as the input approaches a particular point. Here’s a step-by-step guide:

Identify the function f(x) and the point a where you want to find the limit, lim**Step1:**_{x→a}f(x).Set up a table with two columns: one for x values and one for f(x) values.**Step2:**Select values for x that are increasingly close to a from both sides (left and right). For instance, if aaa is 2, choose values like 1.9, 1.99, 2.01, and 2.1.**Step3:**Calculate f(x) for each selected x value.**Step4:**Enter the computed f(x) values in the table next to their corresponding x values.**Step5:**Observe the f(x) values as x gets closer to a. Look for a pattern or a specific value that f(x) approaches.**Step6:**If f(x) approaches a single value L as x approaches a from both sides, then lim**Step7:**_{x→a}f(x) = L. If f(x) approaches different values from different sides, the limit does not exist.

**Example: Find the limit of f(x) = 1/x as x approaches 0.**

** Function:** f(x) = 1/x

** Limit Point:** a = 0

x | f(x) |
---|---|

-0.1 | -10 |

-0.01 | -100 |

0.01 | 100 |

0.1 | 10 |

- Since f(x) approaches −∞ from the left and +∞ from the right, the limit lim
_{x→0}1/x does not exist.

## Methods to Find The Limit of a Function

Methods to Find The Limit of a Function are explained below with the help of example.

### Direct Substitution

Try plugging the value that x is approaching directly into the function.

**Example: Find lim**_{x→2}** (x**^{2}** + 3x – 1)**

**Solution:**

Simply substitute x = 2

= (2)

+ 3(2) – 1^{2}= 4 + 6 – 1

= 9

Therefore, lim

_{x→2}(x+ 3x – 1) = 9^{2}

### Factoring

If direct substitution leads to an indeterminate form like 0/0, try factoring the numerator and/or denominator.

**Example: Find lim**_{x→3}** (x**^{2}** – 9)/(x – 3)**

**Solution:**

Factor numerator: (x + 3)(x – 3) / (x – 3)

Cancel common factors: lim(x→3) (x + 3) = 6

### Rationalization

For limits involving square roots, try rationalizing the numerator or denominator.

**Example: Find lim**_{x→4}** (√x – 2)/(x – 4)**

**Solution:**

Multiply by conjugate: (√x – 2)(√x + 2) / (x – 4)(√x + 2)

Simplify: (x – 4) / ((x – 4)(√x + 2))

Cancel common factors: 1 / (√x + 2)

Now substitute: 1 / (√4 + 2) = 1/4

### Using Special Limit Rules

Some limits have known values or can be evaluated using special rules.

lim | |

lim_{x→∞}(1 + 1/x)^{x} | e |

lim_{x→0}sin(x)/x | 1 |

lim_{x→0}[1 − cos(x)]/x^{2} | 1/2 |

### L’Hôpital’s Rule

If you encounter an indeterminate form like 0/0 or ∞/∞, you can use L’Hôpital’s Rule.

**Example: Find lim**_{x→0}**(1 – cos x)/x**^{2}**.**

**Solution:**

Apply L’Hôpital’s Rule twice:

First application: lim(x→0) (sin x) / (2x)

Second application: lim(x→0) (cos x) / 2 = 1/2

### Squeeze Theorem

If a function is bounded between two functions with the same limit, it must have that limit too.

**Example: Find lim**_{x→0}** x**^{2}** sin(1/x)**

**Solution:**

-|x

| ≤ x^{2}sin(1/x) ≤ |x^{2}|^{2}lim

_{x→0}-|x| = lim(x→0) |x^{2}| = 0^{2}Therefore, lim

_{x→0}xsin(1/x) = 0^{2}

** Read More about **Squeeze Theorem

**.**### Limits at Infinity

For limits as x approaches infinity, divide both numerator and denominator by the highest power of x in the denominator.

**Example: Find lim**_{x→∞}** (3x**^{2}** + 2x – 1)/(x**^{2}** + 5)**

**Solution:**

Divide by x

: lim^{2}_{x→∞}(3 + 2/x – 1/x) / (1 + 5/x^{2})^{2}As x approaches infinity, 1/x and 1/x

approach 0^{2}Therefore, the limit is 3/1 = 3

### Piecewise Functions

For piecewise functions, evaluate the limit using the relevant piece of the function.

**Example: Find lim**_{x→0}** f(x), where **[Tex]f(x) = \begin{cases} x^2 ~\sin(1/x), & \text{if } x ≠ 0,\\ 0, & \text{if } x = 0\end{cases}[/Tex]

**Solution:**

We’ve already shown that lim(x→0) x

^{2}sin(1/x) = 0This matches the function value at x = 0, so the limit exists and equals 0

## Solved Examples: Limit of a Function

**Example 1: Find lim**_{x→2 }**x**^{2}** + 3x – 1.**

**Solution:**

This function is continuous at x = 2, so we can directly substitute x = 2 into the function.

lim

_{x→2}(x^{2}+ 3x – 1) = 2^{2}+ 3(2) – 1 = 4 + 6 – 1 = 9Therefore, lim

_{x→2}(x^{2}+ 3x – 1) = 9

**Example 2: Find lim**_{x→3}** (x**^{2}** – 9)/(x – 3)**

**Solution:**

Direct substitution leads to 0/0, which is indeterminate form.

We need to factor the numerator.

lim

_{x→3}(x^{2}– 9)/(x – 3) = lim_{x→3}[(x+3)(x-3)]/(x – 3)(x-3) terms cancel out:

lim

_{x→3}(x^{2}– 9)/(x – 3) = lim_{x→3}x + 3 = 3 + 3 = 6Therefore, lim

_{x→3}(x^{2}– 9)/(x – 3) = 6

**Example 3: Find lim**_{x→4}**(√x – 2)/(x – 4)**

**Solution:**

Direct substitution leads to 0/0. We’ll rationalize the numerator.

Multiply numerator and denominator by (√x + 2):

lim

_{x→4}[(√x – 2)(√x + 2)] / [(x – 4)(√x + 2)]Simplify the numerator:

lim

_{x→4}(x – 4) / [(x – 4)(√x + 2)]Cancel (x – 4):

lim

_{x→4}1 / (√x + 2)Now we can substitute x = 4:

= 1 / (√4 + 2) = 1/4

Therefore, lim

_{x→4}(√x – 2) / (x – 4) = 1/4

**Example 4: Find lim**_{x→0}** (sin x)/x**

**Solution:**

Direct substitution leads to 0/0. We can apply L’Hôpital’s Rule.

L’Hôpital’s Rule states that for 0/0 or ∞/∞ forms, we can differentiate the numerator and denominator separately.

lim

_{x→0}(sin x) / x = lim_{x→0}(d/dx sin x)/(d/dx x)= lim

_{x→0}cos x / 1Now we can substitute x = 0:

= cos 0 / 1 = 1

Therefore, lim

_{x→0 }(sin x) / x = 1

**Example 5: Find lim**_{x→∞}** (3x**^{2}** + 2x – 1) / (x**^{2}** + 5)**

**Solution:**

For limits at infinity, divide both numerator and denominator by the highest power of x in the denominator (x^2 in this case).

lim

_{x→∞}(3x/x^{2}+ 2x/x^{2}– 1/x^{2}) / (x^{2}/x^{2}+ 5/x^{2})^{2}Simplify:

lim

_{x→∞}(3 + 2/x – 1/x) / (1 + 5/x^{2})^{2}As x approaches infinity, 1/x and 1/x

approach 0:^{2}= 3 / 1 = 3

Therefore, lim

_{x→∞}(3x+ 2x – 1) / (x^{2}+ 5) = 3^{2}These examples demonstrate various techniques for finding limits, including direct substitution, factoring, rationalization, L’Hôpital’s Rule, and handling limits at infinity. The key is to identify the appropriate method based on the function and the point at which the limit is being evaluated.

## Practice Questions

1. Find lim_{x→3} (x** ^{2}** – 9) / (x – 3)

2. Find lim_{x→0} (tan x) / x

3. Find lim_{x→∞} (2x** ^{3}** – x

**+ 5x – 3) / (x**

^{2}**+ 2x – 1)**

^{3}4. Find lim_{x→2} (x** ^{3}** – 8) / (x

**– 4)**

^{2}5. Find lim_{x→0} (1 – cos 2x)/x^{2}

6. Find lim_{x→∞} (x** ^{2}** + 3x + 2)/(2x

**– x + 1)**

^{2}7. Find lim_{x→1} (x** ^{4}** – 1) / (x

**– 1)**

^{2}8. Find lim_{x→0} (sin 3x) / (2x)

9. Find lim_{x→∞}(3^{x}) / (x!)

10. Find lim_{x→0} [ln(1 + x)]/x

## Conclusion

Finding limits involves various techniques such as direct substitution, factoring, rationalization, and using special limit rules or L’Hôpital’s rule for indeterminate forms. The process requires a strong foundation in algebraic manipulation and an understanding of function behavior. Mastering these techniques enables students to analyze function continuity, determine asymptotes, and lay the groundwork for understanding derivatives and integrals in calculus.

## FAQs

### What is the formula for finding the limit of a function?

Here is a statement about limits: “The limit of a function f(x) as x approaches (x-value) is exactly equal to (y-value). We call it limxâ(x-value). Write f( x)=(y-value). For example, limxâ5(x2â2)=23.

### How do you determine if the limit of a function exists?

If both limits are defined for a function at a particular x-value c, and these values coincide, then the limit exists and is equal to the value of the one-sided limit. If the values of the one-sided limits do not coincide, then the two-sided limit does not exist.

### What are the basic rules of limits?

The limit of a difference is equal to the difference of the limits. The limit of a constant times a function is equal to the constant times the limit of the function. The limit of a product is equal to the product of the limits. The limit of a quotient is equal to the quotient of the limits.

### What are the formula of limits?

Limit formula: – Let y = f(x) be a function of x. If at some point x = a f(x) takes on an indefinite form, then we can consider the value of the function very close to a. If these values tend to a particular unique number in the same way that x tends to a, then the resulting unique number is called the limit of f(x) at x = a.

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