Problem 27 - Calculus: Early Transcendentals

Question

Find the following limits or state that they do not exist. Assume \(a, b, c,\) and k are fixed real numbers. $$\lim _{p \rightarrow 2} \frac{3 p}{\sqrt{4 p+1}-1}$$

Step-by-Step Solution

Answer

Question: Determine the limit of the following expression as \(p\) approaches 2: $$\lim _{p \rightarrow 2} \frac{3 p}{\sqrt{4 p+1}-1}$$ Answer: The limit of the given expression as \(p\) approaches 2 is 3.

Step 1: Rationalize the denominator

Multiply both numerator and denominator by the conjugate of the denominator: \(\sqrt{4p+1} + 1\) $$\lim_{p \rightarrow 2} \frac{3 p (\sqrt{4 p + 1} + 1)}{(\sqrt{4 p+1}-1)(\sqrt{4 p+1}+1)}$$ Now, expand the numerator and simplify the denominator (difference of squares).

Step 2: Simplify the expression

Expanding and simplifying the expression, we get: $$\lim_{p \rightarrow 2} \frac{3 p (\sqrt{4 p + 1} + 1)}{(4p + 1) - 1^2}$$ $$\lim_{p \rightarrow 2} \frac{3 p (\sqrt{4p + 1} + 1)}{4p}$$ Now, cancel out the common term \(p\) from the numerator and the denominator.

Step 3: Cancel out the common term

Cancel out the common term and rewrite the expression as: $$\lim_{p \rightarrow 2} \frac{3 (\sqrt{4p + 1} + 1)}{4}$$

Step 4: Find the limit

Plug in the value of \(p\) as it goes to 2: $$\lim_{p \rightarrow 2} \frac{3 (\sqrt{4p + 1} + 1)}{4} = \frac{3 (\sqrt{4(2) + 1} + 1)}{4}$$ $$= \frac{3 (\sqrt{9} + 1)}{4} = \frac{3(3 + 1)}{4} = \frac{12}{4}$$ So, the limit of the given expression as \(p\) goes to 2 is: $$\lim _{p \rightarrow 2} \frac{3 p}{\sqrt{4 p+1}-1} = 3$$

Key Concepts

Rationalization

Rationalization is a powerful algebraic technique often used to simplify expressions involving square roots. In calculus, particularly for limit problems, it proves pivotal in eliminating irrational numbers from denominators. To rationalize an expression, you multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate is the same as the original expression but with the opposite sign between terms.For example, given a denominator like \( \sqrt{4p+1}-1 \), its conjugate is \( \sqrt{4p+1}+1 \). Multiplying the original fraction by this conjugate simplifies the denominator, which helps with further simplification.

• This technique transforms the original irrational denominator into a rational one, making it easier to handle.
• It often results in terms that can be cancelled out, making limit evaluation straightforward.

Rationalization is especially useful when dealing with limits of expressions that involve square roots because it often turns otherwise unmanageable limits into simpler forms.

Difference of Squares

The difference of squares is a fundamental algebraic identity used frequently in simplification problems, especially for rationalizing denominators in limits. It takes the form \( a^2 - b^2 = (a-b)(a+b) \), where \(a\) and \(b\) are any numbers or expressions.When we rationalize an expression in calculus, this identity allows us to instantly simplify the denominator. Consider our example where the denominator after multiplying with its conjugate becomes \((\sqrt{4p+1})^2 - 1^2\). By applying the difference of squares formula, this simplifies to \(4p + 1 - 1\), effectively turning the problematic square root denominator into a simple linear expression, \(4p\).

• This is essential in limit problems as it helps to eliminate the root, which might complicate substitution of values.
• The simplified form often directly unveils terms that can be further reduced, paving the way for solving the limit comfortably.

Limit Evaluation Techniques

Limit evaluation techniques in calculus are strategies used to find the value that a function approaches as the input (a certain variable) approaches a particular point. In the exercise above, we utilized rationalization and the difference of squares to simplify the expression, allowing us to easily find the limit.Once simplification resulted in \( \lim_{p \to 2} \frac{3 (\sqrt{4p + 1} + 1)}{4} \), we used basic substitution techniques to evaluate the limit. Here’s how:- After simplifying, we simply substitute \(p = 2\) into the function to solve for the limit.- This resulted in \( \frac{3 (\sqrt{9} + 1)}{4} \), direct computation yields \( \frac{12}{4} = 3 \).

• Substitution is the simplest limit evaluation technique when the expression is continuous at the point.
• Other techniques, when substitution or simplification fails, include L'Hopital’s Rule or algebraic factoring.

These methods form an essential toolkit for any calculus student as they tackle more complex limit problems. Each technique helps transform challenging limits into solvable problems using clear, logical steps.