Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the limits. \(\lim _{x \rightarrow \infty} \frac{\sqrt{2 x+1}}{x+4}\)

Step-by-Step Solution

Answer

The limit is \( \sqrt{2} \).

Step 1: Analyze the Function

The function given is \( \frac{\sqrt{2x + 1}}{x + 4} \). As \( x \to \infty \), both the numerator and the denominator grow indefinitely, leading to an indeterminate form. We need to simplify this to find the limit.

Step 2: Simplify the Fraction

To simplify, let's divide both the numerator and the denominator by \( x \):\[\frac{\sqrt{2x + 1} / x}{x/x + 4/x} = \frac{\sqrt{\frac{2x + 1}{x}}}{1 + \frac{4}{x}}\]

Step 3: Simplify Further

As \( x \to \infty \), the expression \( \frac{2x + 1}{x} \) simplifies to \( \frac{2x}{x} + \frac{1}{x} = 2 + \frac{1}{x} \). Therefore, \[ \sqrt{\frac{2x + 1}{x}} = \sqrt{2 + \frac{1}{x}} \]

Step 4: Evaluate the Limit

Now, consider the simplified expression: \[ \frac{\sqrt{2 + \frac{1}{x}}}{1 + \frac{4}{x}} \]As \( x \to \infty \), both \( \frac{1}{x} \to 0 \) and \( \frac{4}{x} \to 0 \). Thus, the expression approaches\[ \frac{\sqrt{2}}{1} = \sqrt{2} \]

Step 5: Conclusion

The limit of the original function as \( x \to \infty \) is the value we found: \( \sqrt{2} \). Therefore, \[ \lim_{x \to \infty} \frac{\sqrt{2x + 1}}{x + 4} = \sqrt{2} \]

Key Concepts

Indeterminate Forms

When dealing with limits, especially as a variable approaches infinity, we often encounter expressions that seem undefined, or "indeterminate". An indeterminate form occurs when both the numerator and the denominator of a fraction go to infinity. This makes it difficult to directly compute the limit. For example, \( \frac{\sqrt{2x+1}}{x+4} \) can be considered an indeterminate form as \( x \) approaches infinity, because both \( \sqrt{2x+1} \) and \( x+4 \) grow infinitely large. Indeterminate forms require special techniques to resolve and find the actual limit of the function. Simplification of the original expression is key to overcoming this challenge and accurately evaluating the limit.

Limit Simplification

To tackle indeterminate forms, we routinely simplify the expression. The most common method involves algebraic manipulation that makes the behavior of the function more apparent as the variable approaches infinity. In our problem, \( \frac{\sqrt{2x+1}}{x+4} \), we can simplify the expression by dividing both the numerator and the denominator by \( x \). This technique reduces the expression to:

• Numerator: \( \sqrt{\frac{2x+1}{x}} \) becomes \( \sqrt{2 + \frac{1}{x}} \)
• Denominator: \( \frac{x+4}{x} \) becomes \( 1 + \frac{4}{x} \)

By simplifying the limit to its core components, we can better evaluate the behavior of the function as \( x \to \infty \). Simplification serves as a critical step because it makes the problem more approachable and neatly sets the stage for limit evaluation.

Limit Evaluation

Once the mathematical expression is in its simplest form, evaluating the limit becomes more straightforward. The trick lies in recognizing how terms behave as \( x \) approaches infinity.Taking the simplified expression from our example, \( \frac{\sqrt{2 + \frac{1}{x}}}{1 + \frac{4}{x}} \), evaluate each component. As \( x \to \infty \):

• \( \frac{1}{x} \to 0 \)
• \( \frac{4}{x} \to 0 \)

Thus, the expression simplifies further to \( \frac{\sqrt{2}}{1} \), giving us the evaluated limit \( \sqrt{2} \). The process shows how each term approaches its limiting value, resetting the original equation into a solvable form.

Square Roots in Limits

Square roots in calculus limits can initially complicate the evaluation process. These functions often grow at different rates compared to polynomial expressions, leading to indeterminate forms. Consider the expression \( \sqrt{2x+1} \), which appears in our initial problem.When simplifying such terms, it's important to factor out the highest power of \( x \) that appears under the square root. This allows us to separate it into principal components. For instance, \( \sqrt{2x + 1} = \sqrt{x^2 \cdot \left(2 + \frac{1}{x}\right)} \).Breaking it down per the rules of square roots, \( \sqrt{2x+1} \) simplifies into \( \sqrt{x^2}\cdot\sqrt{2 + \frac{1}{x}} \) or simply \( x \cdot \sqrt{2 + \frac{1}{x}} \).Approaching square roots correctly ensures a deeper and clearer understanding of the limit behavior as it resolves within the larger function.