Problem 27 - Thomas Calculus

Question

The process by which we determine limits of rational functions applies equally well to ratios containing noninteger or negative powers of \(x :\) Divide numerator and denominator by the highest power of \(x\) in the denominator and proceed from there. Find the limits in Exercises \(23-36\). $$\lim _{x \rightarrow \infty} \frac{2 \sqrt{x}+x^{-1}}{3 x-7}$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify the highest power in the denominator

The given function is \( \frac{2 \sqrt{x}+x^{-1}}{3x-7} \). First, identify the highest power of \( x \) in the denominator, which is \( x^1 \) since the term is \( 3x \).

Step 2: Divide each term by the highest power

Divide each term in both the numerator and the denominator by \( x^1 \). This results in \( \frac{\frac{2 \sqrt{x}}{x} + \frac{x^{-1}}{x}}{\frac{3x}{x} - \frac{7}{x}} \).

Step 3: Simplify the terms

Simplify each term: \( \frac{2\sqrt{x}}{x} = 2x^{-1/2} \), \( \frac{x^{-1}}{x} = x^{-2} \), \( \frac{3x}{x} = 3 \), and \( \frac{7}{x} = 7x^{-1} \). The expression becomes \( \frac{2x^{-1/2} + x^{-2}}{3 - 7x^{-1}} \).

Step 4: Evaluate the limit as \( x \rightarrow \infty \)

As \( x \rightarrow \infty \), the terms \( 2x^{-1/2} \), \( x^{-2} \), and \( 7x^{-1} \) all approach 0. Thus, the expression simplifies to \( \frac{0 + 0}{3 - 0} = \frac{0}{3} = 0 \). The limit is 0.

Key Concepts

Understanding Rational Functions

Rational functions are expressions that can be written as a ratio of two polynomials. This means they have the general form \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomial functions. A polynomial function, in simpler terms, is an expression made up of variables and coefficients, that only uses the operations of addition, subtraction, multiplication, and non-negative integer exponents.Rational functions are a significant topic in calculus because of their interesting properties:

• They can have vertical asymptotes, which occur where the denominator is zero.
• They might also have horizontal or oblique asymptotes based on the degrees of the polynomials in the numerator and denominator.
• They can be continuous on their domain, except possibly at points where the denominator is zero.

By understanding rational functions, we can easily solve problems involving their limits, like evaluating how they behave as \( x \) approaches certain values.

Decoding Noninteger Powers

Noninteger powers might seem intimidating at first, but they just refer to exponents that aren't whole numbers. In our problem, for instance, there is the term \( 2\sqrt{x} \), which can also be written as \( 2x^{1/2} \). This is because the square root of a number is equivalent to raising it to the power of one-half. Similarly, \( x^{-1} \) indicates a reciprocal of \( x \), as it equals \( \frac{1}{x} \).Noninteger powers can include both fraction and decimal exponents, and they obey the same rules as integer exponents:

• The product rule: \( x^a \times x^b = x^{a+b} \).
• The quotient rule: \( \frac{x^a}{x^b} = x^{a-b} \), provided \( x eq 0 \).
• The power rule: \( (x^a)^b = x^{ab} \).

Understanding these concepts can make working with such powers easier and make calculus problems, like evaluating limits, more approachable.

Limit Evaluation Simplified

Evaluating limits, especially of rational functions, can be less daunting with some practice. The process involves determining what value a function approaches as the input (usually \( x \)) gets infinitely large or small. For rational functions, this often involves simplifying the function by dividing both the numerator and the denominator by the highest power of \( x \) found in the denominator.Once the division is complete, we closely analyze how each term behaves as \( x \to \infty \) or \( x \to -\infty \). Often times:

• Terms with higher powers in the denominator will approach zero, as fractional exponents become insignificant compared to whole numbers.
• This simplification usually occurs because fractions or terms involving \( x^{-a} \) diminish as \( x \) grows.
• Remaining terms, typically constant terms, dictate the value of the limit.

In our case, after simplification, the complicated-looking expression reduced to a simple \( \frac{0}{3} = 0 \), making limit evaluation a straightforward task.