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Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow 0} \frac{2^{-x}-1}{5^{x}-1} $$

Short Answer

Expert verified
The limit is \( \frac{-\ln(2)}{\ln(5)} \).

Step by step solution

01

Identify Indeterminate Form

The given limit is \( \lim_{x \to 0} \frac{2^{-x} - 1}{5^{x} - 1} \). Substitute \( x = 0 \) in both the numerator and denominator to check the form.- Numerator: \( 2^{0} - 1 = 1 - 1 = 0 \)- Denominator: \( 5^{0} - 1 = 1 - 1 = 0 \)Thus, the limit gives an indeterminate form of \( \frac{0}{0} \). We can use l'Hôpital's Rule to resolve this.
02

Differentiate the Numerator and the Denominator

According to l'Hôpital's Rule, we differentiate the numerator and the denominator separately.- Differentiating the numerator \( f(x) = 2^{-x} - 1 \) gives: \( f'(x) = -\ln(2) \cdot 2^{-x} \)- Differentiating the denominator \( g(x) = 5^{x} - 1 \) gives: \( g'(x) = \ln(5) \cdot 5^{x} \)
03

Apply l'Hôpital's Rule

Substitute the derivatives into the limit according to l'Hôpital's Rule:\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{-\ln(2) \cdot 2^{-x}}{\ln(5) \cdot 5^{x}} \]
04

Evaluate the Limit

Substitute \( x = 0 \) into the expression:\[ \lim_{x \to 0} \frac{-\ln(2) \cdot 2^{-x}}{\ln(5) \cdot 5^{x}} = \frac{-\ln(2) \cdot 2^{0}}{\ln(5) \cdot 5^{0}} = \frac{-\ln(2)}{\ln(5)} \]Thus, the limit evaluates to \( \frac{-\ln(2)}{\ln(5)} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indeterminate Forms
In calculus, an indeterminate form occurs when an expression cannot be directly computed. These forms typically arise during the evaluation of limits. One common scenario is when both the numerator and the denominator of a fraction approach zero simultaneously, resulting in the form \( \frac{0}{0} \). This is called an indeterminate form because it does not convey enough information to determine a specific numerical limit without further analysis.

Indeterminate forms signal that the direct substitution method will not work, and alternative techniques such as L'Hôpital's Rule may be needed. This rule allows us to simplify the evaluation of limits when confronted with such forms.
  • Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), and others like \(0 \times \infty\), \(\infty - \infty\), \(1^\infty\), \(0^0\), and \(\infty^0\).
  • Recognizing these forms is critical in determining the appropriate mathematical approach to solving limit problems.
Understanding how to deal with indeterminate forms is crucial for mastering calculus limits and the application of differentiation.
Calculus Limits
Calculus limits are foundational to understanding changes and behaviors in mathematical functions. A limit describes the value that a function approaches as the input approaches a specific point. Limits help in defining key concepts like continuity, derivatives, and integrals, the three pillars of calculus.

To find a limit, one approach is through direct substitution of the value into the function. However, if direct substitution results in an indeterminate form, other methods such as algebraic simplification or L'Hôpital's Rule must be used. In our problem, the limit of \( \frac{2^{-x} - 1}{5^{x} - 1} \) as \( x \rightarrow 0 \) presented an indeterminate form, leading us to use L'Hôpital's Rule.
  • Limits are not only about finding the exact value but can also be about understanding the trend or behavior of functions as they approach certain values.
  • Mastery of limits includes familiarity with both finite and infinite limits and the various techniques to solve them.
This understanding of limits is critical for further exploring differentiations and integrations in calculus.
Differentiation
Differentiation is an essential concept in calculus that involves finding the derivative of a function, which represents the rate at which the function's value changes with respect to changes in the input.

In the application of L'Hôpital's Rule, differentiation is used to handle indeterminate forms by calculating the derivatives of the function's numerator and denominator separately. The rule states that if an indeterminate form like \( \frac{0}{0} \) is encountered, the limit of the function can be evaluated by differentiating the numerator and denominator until a determinate form is achieved. For example, in our problem, we differentiated the numerator \( 2^{-x} - 1 \) to get \( -\ln(2) \cdot 2^{-x} \) and the denominator \( 5^x - 1 \) to get \( \ln(5) \cdot 5^x \).
  • The derivative of a function gives insights into the slope of the tangent line at any point on the graph of the function.
  • Differentiation is used widely across various fields, including physics for motion analysis and in economics for finding cost and revenue rates.
Thus, differentiation is not only a tool for solving indeterminate forms but also for comprehending broader applications in mathematical analysis.

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Most popular questions from this chapter

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