Limits by Rationalization

Limits are a fundamental concept in calculus providing a way to understand the behavior of the functions as they approach a particular point or infinity. One common method to evaluate limits especially when faced with indeterminate forms is rationalization. This technique is essential for the students as it allows them to simplify complex expressions making it easier to analyze and compute limits.

In this article, we will explore the concept of the limits by rationalization covering necessary topics and providing practical examples, FAQs and resources for further learning.

What are Limits?

In calculus, a limit is a fundamental concept that describes the value that a function approaches as the input approaches a certain value. The Limits help us understand the behavior of the functions at points where they might not be explicitly defined. For example, the limit of the function f(x) as the x approaches a value a is denoted as:

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

This concept is crucial for defining derivatives, integrals, and continuity.

How to Find Limits Using Rationalization

Rationalization is a technique used to simplify expressions involving the roots. It is especially useful when calculating the limits that initially present an indeterminate form like \frac{0}{0}.

The method involves multiplying the numerator and the denominator by the conjugate or an appropriate expression that eliminates the root or simplifies the expression.

Steps to Find Limits Using Rationalization

To find limits, using rationalization; we can use following steps:

Step 1: Identify the Indeterminate Form: If a limit results in an indeterminate form like \frac{0}{0} consider using the rationalization.

Step 2: Multiply by the Conjugate: Multiply the expression by the conjugate of the numerator or the denominator depending on where the root is present.

Step 3: Simplify the Expression: Expand and simplify the expression.

Step 4: Evaluate the Limit: After simplification substitute the limit value and calculate the limit.

Steps to Rationalize

To rationalize rational expressions, we can use following steps:

• Step 1: Identify the Expression.
  • Recognize the limit expression that involves a radical.
• Step 2: Multiply by the Conjugate.
  • Use the conjugate of the expression to multiply the numerator and denominator.
• Step 3: Simplify.
  • After multiplying simplify the expression to eliminate the radical.
• Step 4: Evaluate the Limit.
  • Substitute the limit value into the simplified expression.

Conclusion

The Limits by rationalization is a vital concept in the calculus that aids in understanding the behavior of the functions as they approach specific values. By mastering this technique students can enhance their problem-solving skills and build a solid foundation for the more advanced topics in the calculus. For more information on the limits and related concepts visit the GeeksforGeeks portal for the comprehensive articles and resources.

Read More,

• Limits in calculus
• Determining limits using Algebraic manipulation
• Limits formula
• Limits Practice Problem

Example of Limits by Rationalization

Example 1 : Evaluate \lim_{{x \to 4}} \frac{x - 4}{\sqrt{x} - 2}.

Solution:

The limit initially gives an indeterminate form \frac{0}{0}. To resolve this we rationalize the numerator.

The Multiply numerator and denominator by the conjugate:

\lim_{{x \to 4}} \frac{x - 4}{\sqrt{x} - 2} \times \frac{\sqrt{x} + 2}{\sqrt{x} + 2} = \lim_{{x \to 4}} \frac{(x - 4)(\sqrt{x} + 2)}{(\sqrt{x} - 2)(\sqrt{x} + 2)}

Simplify:

= \lim_{{x \to 4}} \frac{x - 4}{\sqrt{x} - 4} \times \frac{\sqrt{x} + 2}{1} = \lim_{{x \to 4}} \frac{\sqrt{x} + 2}{1} = \frac{4 + 2}{1} = 6

Example 2 : Find \lim_{{x \to 1}} \frac{\sqrt{x + 3} - 2}{x - 1}.

Solution:

The expression gives \frac{0}{0} so we rationalize the numerator:

\lim_{{x \to 1}} \frac{\sqrt{x + 3} - 2}{x - 1} \times \frac{\sqrt{x + 3} + 2}{\sqrt{x + 3} + 2} = \lim_{{x \to 1}} \frac{(x + 3) - 4}{(x - 1)(\sqrt{x + 3} + 2)}

Simplify:

= \lim_{{x \to 1}} \frac{x - 1}{(x - 1)(\sqrt{x + 3} + 2)} = \lim_{{x \to 1}} \frac{1}{\sqrt{x + 3} + 2} = \frac{1}{\sqrt{1 + 3} + 2} = \frac{1}{4}

Example 3 : Evaluate \lim_{{x \to 0}} \frac{\sqrt{1 + x} - 1}{x}.

Solution:

Multiply by the conjugate:

\lim_{{x \to 0}} \frac{\sqrt{1 + x} - 1}{x} \times \frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1} = \lim_{{x \to 0}} \frac{(1 + x) - 1}{x(\sqrt{1 + x} + 1)}

Simplify:

= \lim_{{x \to 0}} \frac{x}{x(\sqrt{1 + x} + 1)} = \lim_{{x \to 0}} \frac{1}{\sqrt{1 + x} + 1} = \frac{1}{2}

Example 4 : Evaluate \lim_{{x \to 9}} \frac{x - 9}{\sqrt{x} - 3}.

Solution:

The expression gives \frac{0}{0}. Rationalize the numerator:

\lim_{{x \to 9}} \frac{x - 9}{\sqrt{x} - 3} \times \frac{\sqrt{x} + 3}{\sqrt{x} + 3} = \lim_{{x \to 9}} \frac{(x - 9)(\sqrt{x} + 3)}{(\sqrt{x} - 3)(\sqrt{x} + 3)}

Simplify:

= \lim_{{x \to 9}} \frac{x - 9}{\sqrt{x} - 9} \times \frac{\sqrt{x} + 3}{1} = \lim_{{x \to 9}} \frac{\sqrt{x} + 3}{1} = \frac{9 + 3}{1} = 12

Example 5 : Evaluate \lim_{{x \to 0}} \frac{\sqrt{4 + x} - 2}{x}.

Solution:

The Multiply by the conjugate:

\lim_{{x \to 0}} \frac{\sqrt{4 + x} - 2}{x} \times \frac{\sqrt{4 + x} + 2}{\sqrt{4 + x} + 2} = \lim_{{x \to 0}} \frac{(4 + x) - 4}{x(\sqrt{4 + x} + 2)}

Simplify:

= \lim_{{x \to 0}} \frac{x}{x(\sqrt{4 + x} + 2)} = \lim_{{x \to 0}} \frac{1}{\sqrt{4 + x} + 2} = \frac{1}{4}

Practice Problems

Problem 1: Evaluate \lim_{{x \to 16}} \frac{\sqrt{x} - 4}{x - 16}.

Problem 2: Find \lim_{{x \to 1}} \frac{\sqrt{2x + 3} - 2}{x - 1}.

Problem 3: Evaluate \lim_{{x \to 0}} \frac{\sqrt{9 + x} - 3}{x}.

Problem 4: Find \lim_{{x \to 4}} \frac{\sqrt{x + 5} - 3}{x - 4}.

Problem 5: Evaluate \lim_{{x \to 25}} \frac{\sqrt{x} - 5}{x - 25}.

Problem 6: Find \lim_{{x \to 2}} \frac{\sqrt{5x + 6} - 4}{x - 2}.

Problem 7: Evaluate \lim_{{x \to 0}} \frac{\sqrt{7 + x} - 7}{x}.

Problem 8: Find \lim_{{x \to 1}} \frac{\sqrt{1 + 3x} - 2}{x - 1}.

Problem 9: Evaluate \lim_{{x \to 9}} \frac{\sqrt{x + 7} - 4}{x - 9}.

Problem 10: Find \lim_{{x \to 0}} \frac{\sqrt{2x + 4} - 4}{x}.