Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Identifying the Challenge

The problem requires us to find the limit of the expression \(\lim _{x \rightarrow \infty}(\sqrt{x^{2}+2x}-x)\) as \(x\) approaches infinity.

Step 2: Rationalization Technique

We apply rationalization to simplify the expression. We multiply and divide by the conjugate to eliminate the square root. The conjugate of \(\sqrt{x^{2}+2x}-x\) is \(\sqrt{x^{2}+2x}+x\). Thus, the expression becomes: \(\frac{(\sqrt{x^{2}+2x}-x)(\sqrt{x^{2}+2x}+x)}{\sqrt{x^{2}+2x}+x}\).

Step 3: Expanding the Numerator

Using the difference of squares formula, we expand the numerator: \((\sqrt{x^{2}+2x})^2 - x^2 = x^2 + 2x - x^2 = 2x\). Thus, the expression simplifies to \(\frac{2x}{\sqrt{x^{2}+2x}+x}\).

Step 4: Simplifying the Expression

Now, the problem is to evaluate \(\frac{2x}{\sqrt{x^{2}+2x}+x}\). We solve by dividing the numerator and denominator by \(x\) to get \(\frac{2}{\frac{\sqrt{x^{2}+2x}}{x}+1}\).

Step 5: Taking the Limit

As \(x\) approaches infinity, \(\frac{\sqrt{x^{2}+2x}}{x}\) tends to 1 because \(\sqrt{x^{2}+2x}\) behaves like \(x\) for very large \(x\). Thus, the expression simplifies to \(\frac{2}{1+1} = 1\).

Step 6: Conclusion

The limit of the original expression is 1. Therefore, \(\lim _{x \rightarrow \infty}(\sqrt{x^{2}+2x} - x) = 1\).

Key Concepts

Rationalization Technique

Rationalization is a technique used to simplify expressions, especially when dealing with limits, by eliminating square roots from the numerator or denominator. In our example, \[\lim _{x \rightarrow \infty}(\sqrt{x^{2}+2x}-x),\] the rationalization process involves multiplying and dividing by the conjugate of the expression containing the square root.

Here's how it works:

• The expression we started with has a square root, \(\sqrt{x^{2}+2x}-x\).
• We find its conjugate, which is \(\sqrt{x^{2}+2x}+x\).
• Multiplying both the numerator and denominator by this conjugate transforms the original expression to \(\frac{(\sqrt{x^{2}+2x}-x)(\sqrt{x^{2}+2x}+x)}{\sqrt{x^{2}+2x}+x}\).
• By doing this, the square root in the numerator gets eliminated, making it easier to find the limit.

Using the conjugate helps manage terms involving square roots, which are particularly tricky when evaluating limits. It's a clever manipulation that works well in calculus problems.

Difference of Squares

The difference of squares is a powerful algebraic identity used to simplify expressions. It states that for any two numbers, \(a\) and \(b\), the identity \((a-b)(a+b) = a^2 - b^2\) holds true.

In our limit problem, we apply this identity to simplify the product in the numerator:

• Let's consider \(a = \sqrt{x^{2}+2x}\) and \(b = x\).
• The numerator becomes \((\sqrt{x^{2}+2x})^2 - x^2\).
• Expanding it using the difference of squares gives us \(x^2 + 2x - x^2\), which simplifies to \(2x\).

This simplification is crucial because it transforms a complex expression into one that's manageable and helps in evaluating the limit. The difference of squares formula efficiently reduces problems involving squares and square roots, providing a pathway to solve them.

Infinity in Limits

Limits involving infinity often require special strategies to evaluate because infinity is not a number in the traditional sense. It's a concept signifying unbounded growth.

In this instance, our task is to find the limit of an expression as \(x\) approaches infinity:

• The original expression \(\lim _{x \rightarrow \infty}(\sqrt{x^{2}+2x} - x)\) becomes the simpler form \(\frac{2x}{\sqrt{x^{2}+2x}+x}\) after rationalization and simplification.
• To evaluate this, we divide every term in both the numerator and the denominator by \(x\), to reflect the behavior of terms as \(x\) becomes extremely large.
• This gives us \(\frac{2}{\frac{\sqrt{x^{2}+2x}}{x}+1}\).

For large \(x\), \(\sqrt{x^2 + 2x} \approx x\), leading \(\frac{\sqrt{x^2+2x}}{x}\) to approach 1. Therefore, our expression simplifies to \(\frac{2}{1 + 1} = 1\).

Understanding infinity in limits is key in calculus, as it allows us to analyze behavior of functions as they grow indefinitely. This example illustrates dividing by the highest power of \(x\) to simplify expressions and accurately determine limits at infinity.