Ch. 7 - Transcendental Functions

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 7 - Transcendental Functions

Problem 7.5.69

Chapter 7, Problem 7.5.69

Theory and Applications
L’Hôpital’s Rule does not help with the limits in Exercises 69–76.
Try it—you just keep on cycling. Find the limits some other way.
69. lim (x → ∞) (√(9x + 1)) / (√(x + 1))

Verified step by step guidance

Identify the limit expression: \(\lim_{x \to \infty} \frac{\sqrt{9x + 1}}{\sqrt{x + 1}}\).

Since both numerator and denominator involve square roots of expressions that grow without bound as \(x \to \infty\), try to simplify the expression by factoring out \(x\) inside the square roots.

Rewrite the numerator and denominator as \(\sqrt{x \left(9 + \frac{1}{x}\right)}\) and \(\sqrt{x \left(1 + \frac{1}{x}\right)}\) respectively.

Use the property of square roots to separate the factors: \(\frac{\sqrt{x} \sqrt{9 + \frac{1}{x}}}{\sqrt{x} \sqrt{1 + \frac{1}{x}}}\).

Cancel \(\sqrt{x}\) from numerator and denominator, then evaluate the limit of the remaining expression \(\frac{\sqrt{9 + \frac{1}{x}}}{\sqrt{1 + \frac{1}{x}}}\) as \(x \to \infty\) by substituting \(\frac{1}{x} \to 0\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Limits at Infinity

Limits at infinity describe the behavior of a function as the input grows without bound. Understanding how functions behave as x approaches infinity helps determine the end behavior and whether the limit converges to a finite value, infinity, or does not exist.

Simplifying Expressions Involving Radicals

When dealing with limits involving square roots, it is often helpful to factor and simplify the expression inside the radicals. This can involve dividing numerator and denominator by the highest power of x to reveal dominant terms and simplify the limit evaluation.

L’Hôpital’s Rule and Its Limitations

L’Hôpital’s Rule is used to evaluate indeterminate forms like 0/0 or ∞/∞ by differentiating numerator and denominator. However, it may fail or cause cyclic behavior if repeated application does not resolve the indeterminate form, requiring alternative methods such as algebraic simplification.

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Related Practice

Textbook Question

7. Order the following functions from slowest growing to fastest growing as x→∞.

a. e^x

b. x^x

c. (ln x)^x

d. e^(x/2)

Textbook Question

Since the hyperbolic functions can be expressed in terms of exponential functions, it is possible to express the inverse hyperbolic functions in terms of logarithms, as shown in the following table.

sinh⁻¹x = ln(x + √(x² + 1)), -∞ < x < ∞

cosh⁻¹x = ln(x + √(x² - 1)), x ≥ 1

tanh⁻¹x = (1/2)ln((1+x)/(1-x)), |x| < 1

sech⁻¹x = ln((1+√(1-x²))/x), 0 < x ≤ 1

csch⁻¹x = ln(1/x + √(1+x²)/|x|), x ≠ 1

coth⁻¹x = (1/2)ln((x+1)/(x-1)), |x| > 1

Use these formulas to express the numbers in Exercises 61–66 in terms of natural logarithms.

63. tanh⁻¹(-1/2)

Textbook Question

In Exercises 139–142, find the length of each curve.

139. y = (1/2)(e^x + e^(−x)) from x = 0 to x = 1.

Textbook Question