Limit of a Function Rules | L’Hôpital’s Rule | Limits at Infinity | Squeeze Theorem | Solved Examples, FAQs

📘 Introduction

In mathematics, a limit helps us understand how a function behaves as the input approaches a particular value. It is a fundamental concept in calculus and plays a crucial role in understanding continuity, derivatives, and integrals.

📐 Definition of a Limit

The limit of a function f(x) as x approaches the value a is written as:

\begin{array}{l} \mathbf{\lim_{x \to a} f(x) = L} \end{array}

This means that as x gets closer to a, the function f(x) approaches the value L.

🔍How to Find the Limit of a Function

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’.

Methods to Find The Limit of a Function

Various methods to find the Limit of a Function are explained below :

Using Table Method :

Using a table to find the limit of a function involves approximating the value of the function as the input approaches a particular point.

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→01/x does not exist.

Direct Substitution

If substituting x=a into the function yields a finite result, the limit is simply the value of the function at that point:.

Example: Find lim_x→2 (x² + 3x – 1)

Solution:

Simply substitute x = 2

= (2)² + 3(2) – 1

= 4 + 6 – 1

= 9

Therefore, lim_x→2 (x² + 3x – 1) = 9

Factoring

When direct substitution gives an indeterminate form like 0/0, use factoring the numerator and/or denominator.

Example: Find lim_x→3 (x² – 9)/(x – 3)

Solution:

Factor numerator:

(x + 3)(x – 3) / (x – 3) = x + 3

Cancel common factors and substituting x=3 into the function yields a finite result

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, we get :

1 / (√x + 2)

Now substitute x = 4

lim_x→4 1/(√x + 2) = 1 / (√4 + 2) = 1/4

Using Special Limit Rules

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

limx→∞(1/x)	0
limx→∞​(1 + 1/x​)x	e
limx→0​sin(x)​/x	1
limx→0​[1 − cos(x)]/x2​	1/2

L’Hôpital’s Rule

If you encounter an indeterminate form like 0/0 or ∞/∞, you can use L’Hôpital’s Rule. If we have an indeterminate form 0/0 or ∞/∞ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit

Example: Find lim_x→0(1 – cos x)/x².

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 x² sin(1/x)

Solution:

-|x²| ≤ x² sin(1/x) ≤ |x²|

lim_x→0 -|x²| = lim(x→0) |x²| = 0

Therefore, lim_x→0 x² sin(1/x) = 0

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→∞ (3x² + 2x – 1)/(x² + 5)

Solution:

Divide by x²: lim_x→∞(3 + 2/x – 1/x²) / (1 + 5/x²)

As x approaches infinity, 1/x and 1/x² approach 0

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 f(x)={x2 sin⁡(1/x),if x≠0,0,if x=0f(x)={x2 sin(1/x),0,​if x=0,if x=0​

Solution:

We’ve already shown that lim(x→0) x² sin(1/x) = 0

This matches the function value at x = 0, so the limit exists and equals 0.

💡 Important Properties of Limits

1. Sum Rule

\begin{array}{l} \mathbf{\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)} \end{array}

2. Product Rule

\begin{array}{l} \mathbf{\lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)} \end{array}

3. Quotient Rule : if lim_⁡x→ag(x)≠0, then

\begin{array}{l} \mathbf{\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}} \end{array}

Solved Examples: Limit of a Function

Example 1: Find lim_x→2 x² + 3x – 1.

Solution:

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

lim_x→2 (x² + 3x – 1) = 2² + 3(2) – 1 = 4 + 6 – 1 = 9

Therefore, lim_x→2 (x² + 3x – 1) = 9

Example 2: Find lim_x→3 (x² – 9)/(x – 3)

Solution:

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

We need to factor the numerator.

lim_x→3 (x² – 9)/(x – 3) = lim_x→3 [(x+3)(x-3)]/(x – 3)

(x-3) terms cancel out:

lim_x→3 (x² – 9)/(x – 3) = lim_x→3 x + 3 = 3 + 3 = 6

Therefore, lim_x→3 (x² – 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 / 1

Now we can substitute x = 0:

= cos 0 / 1 = 1

Therefore, lim_x→0 (sin x) / x = 1

Example 5: Find lim_x→∞ (3x² + 2x – 1) / (x² + 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→∞(3x²/x² + 2x/x² – 1/x²) / (x²/x² + 5/x²)

Simplify:

lim_x→∞(3 + 2/x – 1/x²) / (1 + 5/x²)

As x approaches infinity, 1/x and 1/x² approach 0:

= 3 / 1 = 3

Therefore, lim_x→∞ (3x² + 2x – 1) / (x² + 5) = 3

Practice Questions

1. Find lim_x→3 (x² – 9) / (x – 3)

2. Find lim_x→0 (tan x) / x

3. Find lim_x→∞ (2x³ – x² + 5x – 3) / (x³ + 2x – 1)

4. Find lim_x→2 (x³ – 8) / (x² – 4)

5. Find lim_x→0 (1 – cos 2x)/x²

6. Find lim_x→∞ (x² + 3x + 2)/(2x² – x + 1)

7. Find lim_x→1 (x⁴ – 1) / (x² – 1)

8. Find lim_x→0 (sin 3x) / (2x)

9. Find lim_x→∞(3^x) / (x!)

10. Find lim_x→0 [ln(1 + x)]/x

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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