Problem 25 - Calculus: Early Transcendentals

Question

Evaluate the limit, if it exists. \( \displaystyle \lim_{t \to 0}\frac{\sqrt{1 + t} - \sqrt{1 - t}}{t} \)

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Recognize the Indeterminate Form

The given limit is \( \lim_{t \to 0}\frac{\sqrt{1 + t} - \sqrt{1 - t}}{t} \). As \( t \to 0 \), both the numerator and denominator approach zero, creating an indeterminate form \( \frac{0}{0} \). To resolve this, we can use algebraic manipulation.

Step 2: Multiply by the Conjugate

The term \( \sqrt{1 + t} - \sqrt{1 - t} \) can be simplified by multiplying the expression by its conjugate \( \sqrt{1 + t} + \sqrt{1 - t} \).Multiply both the numerator and the denominator by \( \sqrt{1 + t} + \sqrt{1 - t} \). The limit expression becomes:\[\lim_{t \to 0} \frac{(\sqrt{1 + t} - \sqrt{1 - t})(\sqrt{1 + t} + \sqrt{1 - t})}{t(\sqrt{1 + t} + \sqrt{1 - t})}\]

Step 3: Simplify the Numerator

By multiplying the conjugates, the numerator becomes:\[ (\sqrt{1 + t})^2 - (\sqrt{1 - t})^2 = (1 + t) - (1 - t) = 2t \].The limit expression simplifies to:\[\lim_{t \to 0} \frac{2t}{t(\sqrt{1 + t} + \sqrt{1 - t})}\]

Step 4: Simplify Further and Evaluate the Limit

Cancel the \( t \) in the numerator and denominator:\[\lim_{t \to 0} \frac{2}{\sqrt{1 + t} + \sqrt{1 - t}}\]Now, substitute \( t = 0 \) into the simplified expression:\[\frac{2}{\sqrt{1 + 0} + \sqrt{1 - 0}} = \frac{2}{1 + 1} = \frac{2}{2} = 1\]

Step 5: Conclusion

Since we have no indeterminate form left after simplification and substitution, the limit exists and evaluates to 1.

Key Concepts

Indeterminate Forms

When evaluating limits, you may encounter expressions that seem impossible to determine directly. These are called indeterminate forms. In this problem, as \( t \to 0 \), both the numerator \( \sqrt{1 + t} - \sqrt{1 - t} \) and the denominator \( t \) approach zero, resulting in the indeterminate form \( \frac{0}{0} \). Indeterminate forms mean that you cannot directly compute the limit because both parts of the fraction go to zero. This is often seen in various calculus problems, such as \( 0/0 \), \( \infty - \infty \), and \( 0 \times \infty \). To evaluate these kinds of limits, algebraic manipulation is typically required to simplify the expression into a determinate form that can be calculated without ambiguity.

Conjugate Multiplication

Conjugate multiplication is a technique used to simplify expressions, especially those involving roots or square roots. When we encounter a difference of square roots, such as \( \sqrt{1 + t} - \sqrt{1 - t} \), multiplying by the conjugate \( \sqrt{1 + t} + \sqrt{1 - t} \) facilitates simplification.The conjugate of an expression \( a - b \) is \( a + b \). Multiplying these produces a difference of squares: - \((\sqrt{1 + t} - \sqrt{1 - t})(\sqrt{1 + t} + \sqrt{1 - t}) = (\sqrt{1 + t})^2 - (\sqrt{1 - t})^2 \).This results in full square terms which are easier to manipulate and ultimately simplify the limit analysis. By converting differences into products, it is often easier to cancel terms and reduce complexity.

Limit Simplification

Limit simplification involves reducing a complex limit expression into a simpler form that can be evaluated with basic arithmetic operations. Once we multiply by the conjugate and simplify the numerator, our expression becomes \( \frac{2t}{t(\sqrt{1+t} + \sqrt{1-t})} \). Here, the \( t \) in the numerator and denominator can be cancelled.The simplified expression is \( \frac{2}{\sqrt{1 + t} + \sqrt{1 - t}} \). Now, as \( t \to 0 \), the square roots turn into easily calculable numbers. Substituting \( t = 0 \) yields:- \( \frac{2}{\sqrt{1 + 0} + \sqrt{1 - 0}} = \frac{2}{2} = 1 \).With the complex parts removed, finding the limit becomes straightforward, allowing the identification of the exact value without ambiguity.

Calculus Problems

Calculus is full of intriguing problems, especially when it comes to limits. These problems often require a combination of algebraic tricks and understanding of limits. The challenge lies in recognizing the type of indeterminate form and knowing which technique, such as conjugate multiplication, will simplify the calculation. The exercises help to sharpen critical thinking and problem-solving skills by:

• Encouraging exploration of different approaches to simplify expressions.
• Enhancing algebraic manipulation abilities.
• Building a foundation for more advanced calculus techniques, such as L'Hôpital's rule or series expansion.

With practice and familiarity with these calculus problems, students gain the confidence to tackle more complex scenarios, paving the way for their success in advanced mathematics and scientific applications.