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Magazine Chapter 3: Problem 35
Find the limits. $$\lim _{x \rightarrow+\infty}(\sqrt{x^{2}+x}-x)$$
Short Answer
Expert verified
The limit is \( \frac{1}{2} \).
Step by step solution
01
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.
02
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. \[ \sqrt{x^2 + x} = \sqrt{x^2 (1 + \frac{x}{x^2})} = \sqrt{x^2} \cdot \sqrt{1 + \frac{1}{x}} = x \sqrt{1 + \frac{1}{x}} \]
03
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:\[ x(\sqrt{1+\frac{1}{x}} - 1) \].
04
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:\[ \lim_{x \to +\infty} x \left( 1 + \frac{1}{2x} - 1 \right) = \lim_{x \to +\infty} x \cdot \frac{1}{2x} = \lim_{x \to +\infty} \frac{1}{2} = \frac{1}{2}\]
05
Conclude the result
Thus, the limit as \( x \rightarrow +\infty \) of \( \sqrt{x^{2}+x}-x \) is \( \frac{1}{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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 \( x \).
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:
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.
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 \( x\sqrt{1 + \frac{1}{x}} - x \).
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 \( \lim_{x \to +\infty} x \cdot \frac{1}{2x} = \frac{1}{2} \).
