Problem 35 - Calculus - AP Edition

Question

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

Step-by-Step Solution

Answer

The limit is \( \frac{1}{2} \).

Step 1: Analyze the expression

Our goal is to find the limit as \( x \to +\infty \) for the expression \( \sqrt{x^2 + x} - x \). As \( x \) approaches infinity, both \( \sqrt{x^2 + x} \) and \( x \) become large, so we need to simplify this expression.

Step 2: Factor out the largest power of x

In the expression \( \sqrt{x^2 + x} - x \), notice that the dominant term under the square root is \( x^2 \). We can factor \( x^2 \) out of the square root to help simplify. \[ \sqrt{x^2 + x} = \sqrt{x^2 (1 + \frac{x}{x^2})} = \sqrt{x^2} \cdot \sqrt{1 + \frac{1}{x}} = x \sqrt{1 + \frac{1}{x}} \]

Step 3: Substitute and simplify the expression

Substitute \( x \sqrt{1 + \frac{1}{x}} \) back into the original expression to get \[ x\sqrt{1 + \frac{1}{x}} - x \] Factor x out:\[ x(\sqrt{1+\frac{1}{x}} - 1) \].

Step 4: Find the limit

Now, we need to find the limit of \( x(\sqrt{1+\frac{1}{x}} - 1) \) as \( x \to +\infty \). Notice that for very large \( x \), \( \frac{1}{x} \) becomes very small, and so \( \sqrt{1 + \frac{1}{x}} \approx 1 + \frac{1}{2x} \) using the binomial approximation for small \( z \), where \( \sqrt{1+z} \approx 1 + \frac{z}{2} \). Thus:\[ \lim_{x \to +\infty} x \left( 1 + \frac{1}{2x} - 1 \right) = \lim_{x \to +\infty} x \cdot \frac{1}{2x} = \lim_{x \to +\infty} \frac{1}{2} = \frac{1}{2}\]

Step 5: Conclude the result

Thus, the limit as \( x \rightarrow +\infty \) of \( \sqrt{x^{2}+x}-x \) is \( \frac{1}{2} \).

Key Concepts

Infinity

Infinity in calculus represents a concept rather than a real number. It describes values that grow larger without bound. In our exercise, we are primarily concerned with what happens as a variable approaches infinity.
The expression \( \lim_{x \to +\infty} (\sqrt{x^2 + x} - x) \) shows us how to handle functions as \( x \) becomes very large. Key takeaways include:

• Understanding that as \( x \) approaches infinity, terms like \( x^2 \) dominate over smaller terms like \( x \) or constants.
• Recognizing the importance of simplifying expressions to identify leading behavior, since \( \sqrt{x^2 + x} \) behaves like \( x \) for very large \( x \).

When dealing with infinity, simplifications help us focus on the significant components, like dominant terms, that determine the behavior of the function.

Square Roots in Limits

Square roots can present challenges when computing limits, especially when they are combined with other functions. In our exercise, simplifying \( \sqrt{x^2 + x} - x \) involves understanding how to manage the square root in the limit process.
Here are some highlights:

• Simplifying square roots with variable expressions typically involves factoring out the highest power of \( x \). For example, \( \sqrt{x^2 + x} = \sqrt{x^2(1 + \frac{1}{x})} \).
• This then reduces to \( x\sqrt{1 + \frac{1}{x}} \) by applying the property that \( \sqrt{a^2} = a \) when \( a \) is positive.

By expressing square roots in terms of dominant variables, we gain clarity in evaluating the behavior of the expression as \( x \to +\infty \). This turns a potentially complex problem into a manageable one.

Binomial Approximation

The binomial approximation is a useful technique in calculus to simplify expressions when variables are small. Specifically, for a small\( z \), \( \sqrt{1+z} \approx 1 + \frac{z}{2} \).
This technique helped us evaluate the expression \( x\sqrt{1 + \frac{1}{x}} - x \).

• Recognizing that \( \frac{1}{x} \to 0 \) as \( x \to +\infty \), the binomial approximation allows us to replace \( \sqrt{1 + \frac{1}{x}} \) with \( 1 + \frac{1}{2x} \).
• This simplifies calculations significantly and leads us to the final result of \( \lim_{x \to +\infty} x \cdot \frac{1}{2x} = \frac{1}{2} \).

Using such approximations can transform complex limits into solvable steps, aiding in achieving a final answer without extensive computation.