How to Find the Limit of a Function

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.

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→af(x)=L.

Types of Limits

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

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:

• Step1: Identify the function f(x) and the point a where you want to find the limit, lim⁡_x→af(x).
• Step2: Set up a table with two columns: one for x values and one for f(x) values.
• Step3: 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.
• Step4: Calculate f(x) for each selected x value.
• Step5: Enter the computed f(x) values in the table next to their corresponding x values.
• Step6: Observe the f(x) values as x gets closer to a. Look for a pattern or a specific value that f(x) approaches.
• Step7: If f(x) approaches a single value L as x approaches a from both sides, then lim⁡_x→af(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

• Since f(x) approaches −∞ from the left and +∞ from the right, the limit lim_⁡x→01/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 (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

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

Example: Find lim_x→3 (x² - 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.

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)/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

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→∞ (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) = \begin{cases} x^2 ~\sin(1/x), & \text{if } x ≠ 0,\\ 0, & \text{if } x = 0\end{cases}

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

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

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 (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

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.