Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the limits. \(\lim _{x \rightarrow \infty}\left(\sqrt{2 x^{2}+3}-\sqrt{2 x^{2}-5}\right)\)

Step-by-Step Solution

Answer

0

Step 1: Recognize the Indeterminate Form

The given expression is \ \(\lim_{x \to \infty} \left(\sqrt{2x^2 + 3} - \sqrt{2x^2 - 5}\right)\). As \(x\) approaches infinity, the expression inside the square roots grows without bounds, leading to an indeterminate form of type \(\infty - \infty\). Our goal is to simplify this expression to evaluate the limit.

Step 2: Multiply and Divide by Conjugate

To simplify the expression, we multiply and divide by the conjugate: \[ \lim_{x \to \infty} \frac{\left(\sqrt{2x^2 + 3} - \sqrt{2x^2 - 5}\right) \cdot \left(\sqrt{2x^2 + 3} + \sqrt{2x^2 - 5}\right)}{\sqrt{2x^2 + 3} + \sqrt{2x^2 - 5}} \] This manipulation takes advantage of the identity \((a - b)(a + b) = a^2 - b^2\).

Step 3: Simplify the Numerator

Simplify the numerator using the difference of squares: \[ (\sqrt{2x^2 + 3})^2 - (\sqrt{2x^2 - 5})^2 = (2x^2 + 3) - (2x^2 - 5) = 3 + 5 = 8 \]The numerator simplifies to 8.

Step 4: Simplify the Denominator

Now, simplify the denominator: \[ \sqrt{2x^2 + 3} + \sqrt{2x^2 - 5} \approx 2\sqrt{x^2} = 2x \quad \text{when } x \to \infty \]As \(x\) approaches infinity, higher order terms inside the square roots dominate, approximately equal to \(2x\).

Step 5: Evaluate the Limit

Now that we have a simplified fraction, evaluate the limit:\[ \lim_{x \to \infty} \frac{8}{\sqrt{2x^2 + 3} + \sqrt{2x^2 - 5}} = \lim_{x \to \infty} \frac{8}{2x} = \lim_{x \to \infty} \frac{4}{x} \]Since \(\frac{4}{x}\) approaches 0 as \(x\) approaches infinity, the limit is 0.

Key Concepts

Indeterminate Form

When tackling limits in calculus, you may encounter situations where the form is not easily discernible. These are referred to as *indeterminate forms*. One such form is \( \infty - \infty \). In this scenario, the individual expressions grow infinitely large or small, making direct evaluation challenging.

Recognizing an indeterminate form is the first crucial step in solving limit problems. It signals the need for further manipulation to simplify and evaluate the limit properly. In the expression \( \lim _{x \to \infty}\left(\sqrt{2 x^{2}+3}-\sqrt{2 x^{2}-5}\right) \), as \( x \) heads towards infinity, both square roots tend to infinite values, resulting in an indeterminate form. Our task is to resolve this form to find the limit, and thus, we must employ specific techniques to simplify it.

Conjugate Multiplication

A highly effective strategy for resolving indeterminate forms like \( \infty - \infty \) in expressions involving square roots is *conjugate multiplication*. This involves multiplying and dividing the problematic expression by its conjugate.

The conjugate of \( a - b \) is \( a + b \). When you apply this technique, multiply by the conjugate \( \sqrt{2x^2 + 3} + \sqrt{2x^2 - 5} \). This approach leverages algebraic identities to simplify the problem, eventually allowing us to elucidate the limit.

Why does this work? Well, multiplying by the conjugate exploits the *difference of squares* property. This process crafts a new expression amenable to algebraic simplification, laying the groundwork for resolving the original limit.

Difference of Squares

The difference of squares is a pivotal formula in algebra: \((a - b)(a + b) = a^2 - b^2\). This property simplifies paired terms to reveal new insights. It's particularly useful when simplifying expressions containing square roots.

In our exercise, multiply the expression by its conjugate: \((\sqrt{2x^2 + 3} - \sqrt{2x^2 - 5})(\sqrt{2x^2 + 3} + \sqrt{2x^2 - 5})\). Applying the difference of squares property transforms this expression into:

• \((2x^2 + 3) - (2x^2 - 5)\)
• Simplifying to \(8\)

This transformation effectively eliminates the square roots, allowing you to see the expression more clearly. The result provides a constant numerator of the fraction, which is essential in further reducing and evaluating the limit.

Simplifying Expressions

Simplification is often the last step toward resolving complex calculus problems. Once we've employed strategies like conjugate multiplication and utilized algebraic properties to break down difficult parts, we aim to further simplify the expression to achieve the solution.

Having simplified the numerator to 8, the next step entails approximating the denominator as \( x \to \infty \). The terms \( \sqrt{2x^2 + 3} + \sqrt{2x^2 - 5} \) roughly equal \( 2x \) for large \( x \), because the major terms dominate as \( x \) grows. This simplifies the expression:

• From \( \frac{8}{\sqrt{2x^2 + 3} + \sqrt{2x^2 - 5}} \) to \( \frac{8}{2x} \)
• Thus, simplifies further to \( \frac{4}{x} \)

Taking the limit as \( x \to \infty \), \( \frac{4}{x} \) approaches 0, revealing the solution to our original problem. This simplification illustrates how step-by-step reduction makes solving calculus limits manageable.