Problem 10 - The Calculus with Analytic Geometry

Question

Find the limits, and when applicable indicate the limit theorems being used. $$ \lim _{x \rightarrow+\infty}\left(\sqrt{x^{2}+x}-x\right) $$

Step-by-Step Solution

Answer

\( \frac{1}{2} \)

Step 1: Simplify the expression

Rewrite the term inside the limit in a simpler form. \[ \lim_{{x \to +\infty}} \left( \sqrt{x^2 + x} - x \right) \]

Step 2: Use algebraic manipulation

Multiply and divide the expression by the conjugate to simplify,\[ \lim_{{x \to +\infty}} \left( \sqrt{x^2 + x} - x \right) \cdot \frac{\sqrt{x^2 + x} + x}{\sqrt{x^2 + x} + x} \]

Step 3: Simplify the numerator

Using the conjugate, the numerator becomes: \[ (\sqrt{x^2 + x} - x)(\sqrt{x^2 + x} + x) = (x^2 + x - x^2) = x \]

Step 4: Simplify the denominator

The denominator is: \[ \sqrt{x^2 + x} + x \]

Step 5: Combine the simplified numerator and denominator

Recombine the simplified terms: \[ \lim_{{x \to +\infty}} \frac{x}{\sqrt{x^2 + x} + x} \]

Step 6: Factor out the highest power of x

Factor \( x \) from the numerator and the denominator terms: \[ \lim_{{x \to +\infty}} \frac{x}{x \sqrt{1 + \frac{1}{x}} + x} = \lim_{{x \to +\infty}} \frac{1}{\sqrt{1 + \frac{1}{x}} + 1} \]

Step 7: Take the limit as x approaches infinity

As \( x \to +\infty \), \( \frac{1}{x} \to 0 \). Thus, \[ \lim_{{x \to +\infty}} \frac{1}{\sqrt{1 + \frac{1}{x}} + 1} = \frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{2} \]

Key Concepts

Algebraic Manipulation

In calculus, algebraic manipulation is a powerful tool. It allows us to transform expressions into simpler or more convenient forms. When finding the limit, tweaking the algebraic expressions can reveal hidden patterns or simplify calculations. In this exercise, we initially have the expression \(\frac{\textcolor{blue}{√{x^2 + x} - x}}{\textcolor{red}{1}} \), which seems tricky. To make things manageable, manipulating the expression by multiplying and dividing with a carefully chosen term can simplify the process. This is crucial for making the next steps in limit calculations more straightforward.

Conjugate Method

The conjugate method is especially useful when dealing with square roots in limits. By multiplying the expression by its conjugate, we can eliminate the square root. This technique significantly simplifies the problem. In our exercise, the expression \(\textcolor{red}{√{x^2 + x} - x} \) is tricky because of the square root. By multiplying by the conjugate \(\textcolor{blue}{√{x^2 + x} + x} \), we turn the numerator into something simpler. The steps are as follows:

• Identify the troublesome part, here \(\textcolor{red}{√{x^2 + x} - x} \).
• Multiply and divide by \(\textcolor{blue}{√{x^2 + x} + x} \).
• Combine and simplify the result.

This method helps us handle otherwise complex square root terms.

Asymptotic Behavior

Asymptotic behavior is how functions behave as the input approaches a particular value, in this case, infinity. Understanding this helps predict the behavior of the function for very large or very small values of x. In our example, we study what happens to \(\textcolor{red}{\frac{x}{√{x^2 + x} + x}}\) as \(x \to +\textcolor{blue}{\text{∞}} \).
To grasp this:

• Factor out the highest power of \( x \) in both the numerator and the denominator.
• Examine the remaining terms as \( x \) grows.

We find that \(\textcolor{red}{\frac{1}{√{1 + \frac{1}{x}} + 1}} \) as \( x \) approaches infinity behaves predictably, simplifying our limit calculations.

Infinity Limits

Infinity limits focus on the value a function approaches as the variable heads towards infinity or negative infinity. Understanding these limits is key in calculus. Different techniques are applied based on the problem. For \( x \to + \textcolor{blue}{\text{∞}} \), it's critical to understand that very large values simplify our expressions.
In this exercise, as \( x \to \textcolor{blue}{+\text{∞}} \):

• \(\frac{1}{x} \) approaches \(\textcolor{red}{0} \).
• The expression \(\frac{1}{√{1 + \frac{1}{x}} + 1} \) simplifies as \( x \to + \textcolor{blue}{\text{∞}} \).

Recognizing these behaviors allows for the straightforward calculation of limits involving infinity, leading us to conclude that our original limit is \(\frac{1}{2} \).