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Magazine Chapter 5: Problem 49
Find the limits in Problems 1-60; not all limits require use of l'H么pital's rule. $$ \lim _{x \rightarrow \infty}\left(\frac{x}{1+x}\right)^{x} $$
Short Answer
Expert verified
The limit is 0.
Step by step solution
01
Recognize the Indeterminate Form
The expression to evaluate is \( \lim_{x \rightarrow \infty} \left( \frac{x}{1+x} \right)^x \). Substitute \( x = \infty \) directly to see if it's an indeterminate form. As \( x \rightarrow \infty \), \( \frac{x}{1+x} \rightarrow 1 \), suggesting an indeterminate form \( 1^\infty \).
02
Use the Exponential and Logarithm Transformation
To simplify the indeterminate form, use the natural logarithm to transform the expression. Let \( y = \left( \frac{x}{1+x} \right)^x \), then take \( \ln(y) = x \cdot \ln \left( \frac{x}{1+x} \right) \). Our goal is now to find \( \lim_{x \rightarrow \infty} \ln(y) \).
03
Simplify the Logarithmic Term
Simplify the term \( \ln \left( \frac{x}{1+x} \right) = \ln(x) - \ln(1+x) \). This can be further simplified using the expansion \( \ln(1+x) \approx x \) for large \( x \), so as \( x \rightarrow \infty \), \( \ln(1+x) \approx \ln(x + x) = \ln(2x) \).
04
Combine and Simplify the Limit Process
Combine the approximations and take the limit: \( \ln(y) = x (\ln(x) - \ln(1+x)) \approx x \left( \ln(x) - \ln(2x) \right) = x (\ln(x) - (\ln(2) + \ln(x))) = x (-\ln(2)) \).Thus, \( \ln(y) = -x \ln(2) \) and \( \lim_{x \rightarrow \infty} \ln(y) = -\infty \).
05
Exponentiate to Find Limit
Since \( \ln(y) \rightarrow -\infty \), this implies that \( y = e^{\ln(y)} = e^{-x \ln(2)} \rightarrow e^{-\infty} = 0 \) as \( x \rightarrow \infty \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limits
Limits are a fundamental concept in calculus. They describe how a function behaves as the input approaches a specific value. When we calculate the limit of a function, we are trying to understand the value that the function approaches as the input nears a certain point.
In the original exercise, we're dealing with a limit as the input, called \( x \), approaches infinity. This is an essential skill for analyzing the long-term behavior of functions in calculus.
- Approaching a value: The limit tells us what value a function gets closer to as the input gets closer to a certain point.
- Infinite limits: Sometimes, the input doesn't approach a regular number but instead approaches infinity. Limits at infinity describe the behavior of functions as they grow very large.
In the original exercise, we're dealing with a limit as the input, called \( x \), approaches infinity. This is an essential skill for analyzing the long-term behavior of functions in calculus.
Indeterminate forms
Indeterminate forms occur when a mathematical expression is not clearly defined or doesn't resolve to a particular number upon substitution. Common indeterminate forms include \( \frac{0}{0} \), \( \infty - \infty \), and \( 1^\infty \), to name a few.
In the exercise, the expression \( \left( \frac{x}{1+x} \right)^x \) approaches an indeterminate form of \( 1^\infty \) when \( x \) tends to infinity.
In our problem, instead of using l'H么pital's Rule, we apply a logarithmic transformation to manage the \( 1^\infty \) form effectively.
In the exercise, the expression \( \left( \frac{x}{1+x} \right)^x \) approaches an indeterminate form of \( 1^\infty \) when \( x \) tends to infinity.
- Recognition: Identifying these forms is crucial because it tells us that additional steps are needed to properly evaluate the limit.
- Resolution methods: Approaches like l'H么pital's Rule, transformations using logarithms, or algebraic simplification can resolve these forms.
In our problem, instead of using l'H么pital's Rule, we apply a logarithmic transformation to manage the \( 1^\infty \) form effectively.
Exponential functions
Exponential functions involve expressions where a constant base is raised to a variable exponent, like \( a^x \). These functions are pervasive in calculus and natural sciences due to their tendency to model growth and decay processes.
In our problem, we observe an exponential form \( \left( \frac{x}{1+x} \right)^x \). By transforming it, we use its properties to find its limit as \( x \) goes to infinity, ultimately converting it back using exponential and natural logarithm techniques.
- Growth and decay: Exponential functions are characterized by rapid increase (growth) or decrease (decay).
- Base \( e \): The base \( e \) is special since functions of the form \( e^x \) have unique properties, particularly when differentiating and integrating.
In our problem, we observe an exponential form \( \left( \frac{x}{1+x} \right)^x \). By transforming it, we use its properties to find its limit as \( x \) goes to infinity, ultimately converting it back using exponential and natural logarithm techniques.
Natural logarithm
The natural logarithm, denoted as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \) is approximately equal to 2.71828. It is a vital tool in calculus due to its intricate relationship with the exponential function.
In the exercise, we employed the natural logarithm to transform the power expression into a more manageable form. By taking \( \ln(y) = x \cdot \ln \left( \frac{x}{1+x} \right) \) and simplifying, we could easily find the limit, demonstrating the powerful utility of the natural logarithm in resolving complex expressions.
- Inverse relation: The natural logarithm is the inverse function of the exponential function \( e^x \).
- Calculation simplification: Using \( \ln(x) \) can simplify the calculation of limits that involve exponentials and products.
In the exercise, we employed the natural logarithm to transform the power expression into a more manageable form. By taking \( \ln(y) = x \cdot \ln \left( \frac{x}{1+x} \right) \) and simplifying, we could easily find the limit, demonstrating the powerful utility of the natural logarithm in resolving complex expressions.
