Problem 25 - Calculus: Early Transcendentals

Question

Find the limit or show that it does not exist. \( \displaystyle \lim_{x \to \infty}\frac{\sqrt{x + 3x^2}}{4x - 1} \)

Step-by-Step Solution

Answer

The limit is \(\frac{\sqrt{3}}{4}\).

Step 1: Rewrite the expression

First, we take a look at the expression \( \frac{\sqrt{x + 3x^2}}{4x - 1} \). Notice that within the square root, the term \(3x^2\) is going to dominate since we are evaluating the limit as \(x \to \infty\). This suggests we should factor \(x^2\) out of the square root numerator for simplification.

Step 2: Simplify the square root

Factor \(x^2\) out of the terms inside the square root in the numerator: \( \sqrt{x + 3x^2} = \sqrt{x^2(3 + \frac{1}{x})} \). This simplifies to \( x\sqrt{3 + \frac{1}{x}} \).

Step 3: Simplify the entire fraction

Substitute the simplified form of the square root back into the entire expression: \( \frac{x\sqrt{3 + \frac{1}{x}}}{4x - 1} \). Now, express the denominator as \( 4x(1 - \frac{1}{4x}) \) for easier manipulation.

Step 4: Factor out x from the fraction

Because the terms in the expression \(x\sqrt{3 + \frac{1}{x}}\) and \(4x(1 - \frac{1}{4x})\) both involve \(x\), we can factor \(x\) out, resulting in: \( \frac{\sqrt{3 + \frac{1}{x}}}{4(1 - \frac{1}{4x})} \).

Step 5: Take the limit as x approaches infinity

As \(x\) approaches infinity, the term \(\frac{1}{x}\) approaches zero, so the expression simplifies to \(\frac{\sqrt{3}}{4} \).

Step 6: Conclusion

Therefore, the limit of the given function as \(x\) approaches infinity is \(\frac{\sqrt{3}}{4} \).

Key Concepts

Square Root Simplification

Simplifying square roots can make complex-looking expressions much more manageable, especially when working with limits. In calculus, when you encounter a square root in an expression, it's often helpful to identify and factor out the dominant term. This is key, particularly if you are finding limits at infinity.
\[ \text{If you have an expression like} \ \sqrt{x + 3x^2}, \text{ the term } 3x^2 \text{ dominates as } x \to \infty . \]
Factor the dominant term out:\[ \sqrt{x + 3x^2} = \sqrt{x^2(3 + \frac{1}{x})} = x\sqrt{3 + \frac{1}{x}} .\]
This step leverages the rule \( \sqrt{a^2b} = a\sqrt{b} \). Now, it’s easier to handle the square root and use it in further calculations.

Rational Functions

A rational function is a ratio of two polynomials. In expressions involving limits, simplifying to a clearer form helps to see the behavior of the function as it approaches infinity.

In the expression \( \frac{x\sqrt{3 + \frac{1}{x}}}{4x - 1} \), we deal with such a rational function. By recognizing the leading term in the numerator and the denominator, adjustments can be made to see how they simplify as \(x\) grows very large.

• Recognize and simplify any roots or factors that can be cancelled or balanced.
• The fraction can be further simplified by factoring out common terms like \(x\), producing \( \frac{\sqrt{3 + \frac{1}{x}}}{4 - \frac{1}{4x}} \).

Through these steps, the function gets easier to evaluate at infinity.

Limit at Infinity

Taking the limit of a function as \(x\) approaches infinity reveals how the function behaves at very large values of \(x\). In many cases, part of the process involves determining which terms become negligible.

As seen in \( \frac{\sqrt{3 + \frac{1}{x}}}{4(1 - \frac{1}{4x})} \), the term \(\frac{1}{x}\) becomes negligible as \(x\) grows, simplifying to \(\frac{\sqrt{3}}{4}\).

• Terms like \(\frac{1}{x}\) vanish because they approach zero.
• Focus on dominant terms which actually determine the function's behavior as \(x\) approaches infinity.

By using these strategies, calculating limits at infinity becomes straightforward, returning consistent and meaningful results.