Problem 35 - Thomas Calculus in SI Units

Question

Use l'Hôpital's rule to find the limits. $$\lim _{y \rightarrow 0} \frac{\sqrt{5 y+25}-5}{y}$$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{2} \).

Step 1: Substitute the limit value

First, substitute the value of \( y \) approaching 0 into the expression \( \lim_{y \to 0} \frac{\sqrt{5y + 25} - 5}{y} \). We get the form \( \frac{0}{0} \), which is indeterminate and appropriate for l'Hôpital's rule.

Step 2: Apply l'Hôpital's Rule

Since we have an indeterminate form \( \frac{0}{0} \), we can apply l'Hôpital's Rule. According to the rule, we take the derivative of both the numerator and the denominator. First, find the derivative of the numerator \( \sqrt{5y + 25} - 5 \), which is \( \frac{d}{dy}(\sqrt{5y + 25}) = \frac{1}{2\sqrt{5y+25}} \cdot 5 = \frac{5}{2\sqrt{5y+25}} \). The derivative of the constant \(-5\) is 0. Next, find the derivative of the denominator \( y \), which is 1.

Step 3: Compute the new limit

After taking the derivatives, the expression becomes \( \lim_{y \to 0} \frac{\frac{5}{2\sqrt{5y+25}}}{1} = \lim_{y \to 0} \frac{5}{2\sqrt{5y+25}} \). Now, substitute \( y = 0 \) into this expression: \( \frac{5}{2\sqrt{5(0)+25}} = \frac{5}{2 \times 5} = \frac{5}{10} = \frac{1}{2} \).

Key Concepts

Indeterminate Forms

In calculus, when evaluating the limit of a function as it approaches a certain value, we may encounter expressions that are not straightforward to solve. These are known as "indeterminate forms." When you plug in the limit value, if you obtain forms like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), or similar, these are signals of indeterminate forms.
These forms arise because the behaviour of the function near the limit point is not clearly defined. For example, if both the numerator and the denominator approach zero, it's unclear what the final value of the expression should be without further analysis.

• Common indeterminate forms: \(0/0\), \(\infty/\infty\), \(\infty - \infty\)
• L'Hôpital's Rule can often be applied to resolve these forms

Understanding indeterminate forms is crucial as it guides us on whether we can apply certain rules or need to manipulate the expression further to find a conclusive limit.

Differentiation

Differentiation is the process of finding the derivative of a function. A derivative represents how a function changes as its input changes. It's akin to finding the slope of a function at any given point.
To apply l'Hôpital's Rule effectively, you need to differentiate both the numerator and the denominator of a given function. This involves finding how each part of the fraction changes with respect to the variable of interest.

• The derivative of a constant is always 0. For example, the derivative of \(-5\) is 0.
• For more complex functions, such as \(\sqrt{5y+25}\), we use the chain rule to differentiate. \(\frac{d}{dy}(\sqrt{5y+25}) = \frac{5}{2\sqrt{5y+25}}\).

Differentiation helps simplify complex expressions into forms that are easier to evaluate as a limit, especially when dealing with indeterminate forms.

Limits in Calculus

Limits in calculus are foundational for understanding how functions behave as they approach particular values. Limits help determine what value a function heads towards as the inputs get close to a specific point.
The expression \( \lim_{y \rightarrow 0} \frac{\sqrt{5y+25}-5}{y} \) requires an understanding of limits to solve using l'Hôpital's Rule, especially since direct substitution results in an indeterminate form.

• Limits describe the behavior of a function as it nears a certain point, even if it's not defined at that point.
• In this problem, after applying differentiation, the limit gives \( \frac{1}{2} \) as \(y\) approaches 0.

Understanding limits is essential in calculus as they bridge gaps between discrete values and facilitate the precise calculation of instantaneous rates of change, helping to analyze continuity and behavior of functions.