Problem 58 - Calculus Early Transcendentals: Pearson New International Edition

Question

Begin by plotting the function in an appropriate window. \(\lim _{x \rightarrow-\infty} \sqrt{\frac{2 x^{2}-3 x}{5 x^{2}+1}}\)

Step-by-Step Solution

Answer

The limit is \( \frac{\sqrt{10}}{5} \).

Step 1: Understand the Problem

The problem asks for the limit of the function \( f(x) = \sqrt{\frac{2x^2 - 3x}{5x^2 + 1}} \) as \( x \to -\infty \). This means we need to determine the behavior of this function when \( x \) approaches very large negative values.

Step 2: Simplify the Expression

Observe the dominant terms: in the numerator \(2x^2\) and in the denominator \(5x^2\) as \( x \to -\infty \). Factor these largest terms:\[ \frac{2x^2 - 3x}{5x^2 + 1} = \frac{x^2(2 - \frac{3}{x})}{x^2(5 + \frac{1}{x^2})}. \] Simplify this to \( \frac{2 - \frac{3}{x}}{5 + \frac{1}{x^2}} \).

Step 3: Take the Limit

Compute the limit as \( x \to -\infty \):\[ \lim_{x \to -\infty} \frac{2 - \frac{3}{x}}{5 + \frac{1}{x^2}}. \] As \( x \to -\infty \), fractions with \( x \) in the denominator become zero: \( \frac{3}{x} \to 0 \) and \( \frac{1}{x^2} \to 0 \). Thus,\[ \lim_{x \to -\infty} \frac{2 - \frac{3}{x}}{5 + \frac{1}{x^2}} = \frac{2}{5}. \]

Step 4: Consider the Square Root

Recall we seek \( \lim_{x \to -\infty} \sqrt{\frac{2x^2 - 3x}{5x^2 + 1}} \). From the previous step, we found that \( \lim_{x \to -\infty} \frac{2x^2 - 3x}{5x^2 + 1} = \frac{2}{5} \). Since square roots preserve the sign of the component,\[ \sqrt{\frac{2}{5}} = \frac{\sqrt{10}}{5}. \] Thus, the limit is \( \frac{\sqrt{10}}{5} \).

Key Concepts

Limits

The concept of limits is fundamental in calculus. A limit helps us understand the behavior of a function as it approaches a particular point, which may either be a specific number or infinity. In our exercise, we are interested in finding out what happens to the function \(\lim _{x \rightarrow-\infty} \sqrt{\frac{2 x^{2}-3 x}{5 x^{2}+1}}\)as \(x\) approaches negative infinity.
This essentially asks us to predict what value the function gets closer to when \(x\) becomes very large in the negative direction:

• We look primarily at the dominant terms, which are the highest powers of \(x\) in the numerator and the denominator.
• By simplifying, we can neglect smaller terms, as these effectively become negligible when compared to the larger terms at extreme values of \(x\).
• This simplification allows us to see that the behavior of our function is similar to a simpler function, which is easier to analyze.

When computed, the limit in our exercise resulted in \(\frac{2}{5}\), indicating the direction toward which the function moves as \(x\) becomes indefinitely large and negative.

Square Roots

Understanding square roots is crucial, especially when they appear in functions like the one in our exercise.
A square root function, denoted as \(\sqrt{...}\), essentially gives us a number which, when multiplied by itself, results in the original number under the square root:

• It is always non-negative, which means it only outputs positive values or zero.
• Square roots play a particular role because they influence the final result's sign and magnitude, especially when part of a limit or rational function.
• In the specific context of the exercise, taking the square root of \(\frac{2}{5}\) helps us find the overall limit since the square root operation preserves the non-negative nature of the expression.

Thus, by computing the square root, we refined the solution to \(\frac{\sqrt{10}}{5}\), keeping the mathematical behavior consistent as we execute more complex operations.

Rational Functions

Rational functions, which are quotients of polynomials, are a type of function that often appear in calculus exercises like this one.
They have distinctive characteristics that influence their behavior:

• They are expressed as \(\frac{P(x)}{Q(x)}\), where both \(P\) and \(Q\) are polynomials.
• The behavior of rational functions at extremes (like approaching infinity) is largely influenced by the leading terms. This means the highest power of \(x\) in both the numerator and denominator are most significant.
• Simplifying rational functions is easier when common factors are canceled, resulting in simpler forms to calculate limits.

In our solution, the rational function was simplified to \(\frac{2 - \frac{3}{x}}{5 + \frac{1}{x^2}}\), allowing us to understand its asymptotic behavior. This enables us to determine critical characteristics like approaching values as \(x\) tends towards infinity.