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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.5.37

Use l’Hôpital’s rule to find the limits in Exercises 7–52.
37. lim (y → 0) (√(5y + 25) - 5) / y

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First, identify the form of the limit as \( y \to 0 \) for the expression \( \frac{\sqrt{5y + 25} - 5}{y} \). Substitute \( y = 0 \) to check if it results in an indeterminate form.
Since substituting \( y = 0 \) gives \( \frac{\sqrt{25} - 5}{0} = \frac{5 - 5}{0} = \frac{0}{0} \), which is an indeterminate form, we can apply l'Hôpital's Rule.
Apply l'Hôpital's Rule by differentiating the numerator and denominator separately with respect to \( y \). The numerator is \( \sqrt{5y + 25} - 5 \), so find its derivative:
The derivative of the numerator is \( \frac{d}{dy} \left( \sqrt{5y + 25} - 5 \right) = \frac{1}{2\sqrt{5y + 25}} \times 5 = \frac{5}{2\sqrt{5y + 25}} \). The derivative of the denominator \( y \) is 1.
Rewrite the limit using these derivatives: \( \lim_{y \to 0} \frac{5}{2\sqrt{5y + 25}} \). Then, evaluate this new limit by substituting \( y = 0 \).

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Key Concepts

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

Limits and Indeterminate Forms

Limits describe the behavior of a function as the input approaches a particular value. When direct substitution results in expressions like 0/0 or ∞/∞, these are called indeterminate forms, requiring special techniques to evaluate the limit.
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l’Hôpital’s Rule

l’Hôpital’s Rule is a method for evaluating limits that yield indeterminate forms 0/0 or ∞/∞ by differentiating the numerator and denominator separately and then taking the limit of their quotient.
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Differentiation of Composite Functions

To apply l’Hôpital’s Rule, one must differentiate functions, including composite ones like square roots. This involves using the chain rule, which differentiates the outer function and multiplies by the derivative of the inner function.
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