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Magazine Chapter 3: Problem 13
Evaluate the limit. If the limit is of an indeterminate form, indicate the form and use L'H么pital's Rule to evaluate the limit. $$ \lim _{x \rightarrow 5} \frac{(x-1)^{0.5}-2}{x^{2}-25} $$
Short Answer
Expert verified
The limit is \(\frac{1}{40}\).
Step by step solution
01
Identify the Indeterminate Form
First, we substitute \(x = 5\) into the function to determine if it results in an indeterminate form. Substituting gives \((5-1)^{0.5} - 2 = 2 - 2 = 0\) for the numerator and \(5^2 - 25 = 25 - 25 = 0\) for the denominator. Thus, the expression is of the form \(\frac{0}{0}\), which is indeterminate.
02
Apply L'H么pital's Rule
Since the limit is in the \(\frac{0}{0}\) form, we can apply L'H么pital's Rule. According to this rule, we differentiate the numerator and denominator separately.
03
Differentiate the Numerator
The numerator is \((x-1)^{0.5} - 2\). The derivative of \((x-1)^{0.5}\) is \(\frac{1}{2}(x-1)^{-0.5}\). So, the derivative of the numerator is \(\frac{1}{2}(x-1)^{-0.5}\).
04
Differentiate the Denominator
The denominator is \(x^2 - 25\). The derivative of \(x^2\) is \(2x\) and the derivative of \(-25\) is \(0\). Thus, the derivative of the denominator is \(2x\).
05
Evaluate the Limit of the New Expression
Substitute \(x = 5\) into the expression obtained by applying L'H么pital's Rule: \[\lim _{x \rightarrow 5} \frac{\frac{1}{2}(x-1)^{-0.5}}{2x}.\]At \(x=5\), this becomes \[\frac{\frac{1}{2}(5-1)^{-0.5}}{2\times5} = \frac{\frac{1}{2} \times \frac{1}{2}}{10} = \frac{\frac{1}{4}}{10} = \frac{1}{40}.\]
06
Conclusion
The original limit evaluates to \(\frac{1}{40}\) after using L'H么pital's Rule.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Indeterminate Forms
Indeterminate forms are expressions in calculus that do not have a clear value or limit. When attempting to evaluate a limit, you might encounter expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These are called indeterminate because simply substituting values gives a result that doesn't provide meaningful information about the behavior of the function.
Identifying indeterminate forms is crucial when applying L'H么pital's Rule, which specifically addresses these cases. The exercise at hand is an excellent example: substituting \(x = 5\) into the function results in a \(\frac{0}{0}\) form. Recognizing such forms allows us to move to techniques like L'H么pital's Rule, which helps find the true limit in these tricky situations.
Identifying indeterminate forms is crucial when applying L'H么pital's Rule, which specifically addresses these cases. The exercise at hand is an excellent example: substituting \(x = 5\) into the function results in a \(\frac{0}{0}\) form. Recognizing such forms allows us to move to techniques like L'H么pital's Rule, which helps find the true limit in these tricky situations.
The Process of Limit Evaluation
Limit evaluation is a fundamental concept in calculus used to understand the behavior of functions as they approach a specific point. In this exercise, our goal was to determine the limit of a complicated expression as \(x\) approaches 5.
A direct substitution of \(x = 5\) gave us an indeterminate form, \(\frac{0}{0}\). This is a signal that it鈥檚 time to delve deeper using calculus techniques such as L'H么pital's Rule. Limit evaluation becomes an exploration of the true nature of functions, requiring a nuanced approach beyond simple arithmetic operations.
A direct substitution of \(x = 5\) gave us an indeterminate form, \(\frac{0}{0}\). This is a signal that it鈥檚 time to delve deeper using calculus techniques such as L'H么pital's Rule. Limit evaluation becomes an exploration of the true nature of functions, requiring a nuanced approach beyond simple arithmetic operations.
Differentiation Steps in Applying L'H么pital's Rule
Differentiation is the process of finding the derivative of a function and is a powerful tool in calculus, especially when dealing with L'H么pital's Rule. In the exercise, we start by differentiating the numerator \((x-1)^{0.5} - 2\) to get \(\frac{1}{2}(x-1)^{-0.5}\). This involves recognizing the power rule and chain rule for derivatives.
Next, we differentiate the denominator \(x^2 - 25\), which is a straightforward application of the power rule, resulting in \(2x\). Differentiation allows us to transform the original expression into a form that can provide a definite limit, by eliminating the indeterminate form.
Next, we differentiate the denominator \(x^2 - 25\), which is a straightforward application of the power rule, resulting in \(2x\). Differentiation allows us to transform the original expression into a form that can provide a definite limit, by eliminating the indeterminate form.
Integrating Calculus Techniques
Calculus is the mathematical study of change, and integrating its techniques can help solve complex limit problems effectively. Applying L'H么pital's Rule is a perfect example of leveraging differential calculus to find limits that are otherwise indeterminate.
This exercise showed us the practical use of L'H么pital's Rule as it allowed for the reduction of an indeterminate form to a determinate one. By differentiating the numerator and denominator separately, we turned a difficult \(\frac{0}{0}\) situation into a solvable problem, yielding a final limit of \(\frac{1}{40}\). Such techniques highlight the elegance and utility of calculus in understanding and solving real-world mathematical problems.
This exercise showed us the practical use of L'H么pital's Rule as it allowed for the reduction of an indeterminate form to a determinate one. By differentiating the numerator and denominator separately, we turned a difficult \(\frac{0}{0}\) situation into a solvable problem, yielding a final limit of \(\frac{1}{40}\). Such techniques highlight the elegance and utility of calculus in understanding and solving real-world mathematical problems.
