Chapter 9: Problem 57
Use the fact that \(\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow 0^{+}} f\left(\frac{1}{x}\right)\) to find the limits. $$ \lim _{n \rightarrow \infty}\left(\frac{n-1}{n+1}\right)^{n} $$
Short Answer
Expert verified
The limit is \( e^{-2} \).
Step by step solution
01
Understanding the Limit Transformation
To find \( \lim_{n \rightarrow \infty}\left(\frac{n-1}{n+1}\right)^n \), we first rewrite \( \left(\frac{n-1}{n+1}\right)^n \) using the transformation given in the problem: \( \lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow 0^+} f\left(\frac{1}{x}\right) \). This indicates a transformation involving a change from large \( n \) to small values through inversion.
02
Simplifying the Expression
Observe that \( \frac{n-1}{n+1} \approx 1 - \frac{2}{n+1} \) as \( n \) becomes large. Thus, we have the base \( 1 - \frac{2}{n+1} \). We will be evaluating \( \left(1 - \frac{2}{n+1}\right)^n \).
03
Using the Exponential Limit Approximation
For very small \( x \), \( (1 - x)^n \) can be approximated using the limit \( \ln(1-x) \approx -x \) for small \( x \). Therefore, we have: \((1-x)^n \approx e^{-nx}\). Substitute \( x = \frac{2}{n+1} \): \(\left(1-\frac{2}{n+1}\right)^n \approx e^{-n \cdot \frac{2}{n+1}}.\) As \( n \rightarrow \infty \), it simplifies to: \( e^{-2}.\)
04
Evaluate the Limit
Therefore, the original limit \( \lim_{n \rightarrow \infty}\left(\frac{n-1}{n+1}\right)^n \) is equivalent to evaluating: \( \lim_{n \rightarrow \infty} e^{-2} = e^{-2}. \) As there are no variables left in the expression, this is simply the value \( e^{-2}. \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit Transformation
In calculus, transforming limits is a valuable tool, especially when dealing with expressions that progress towards infinity. One key method involves utilizing the identity \( \lim_{x \to \infty} f(x) = \lim_{x \to 0^+} f\left(\frac{1}{x}\right) \). This transformation switches the behavior of a function at infinity to its behavior as the variable approaches zero. This allows us to handle complex limits by transforming them into more manageable forms.
- Limit transformations are particularly helpful for studying the asymptotic behavior of functions.
- This approach works by essentially flipping the variable, providing new insights by studying the limit from a different perspective.
Exponential Limit Approximation
The exponential limit approximation is a powerful technique when dealing with limits of expressions raised to a power that tend towards zero. When a term like \((1-x)^n\) is present and \(x\) is small, the expression can be approximated with exponentials using the natural logarithm: \( \ln(1-x) \approx -x \) for small \(x\). This allows us to use the famous approximation: \((1-x)^n \approx e^{-nx}\).
Using this approximation makes complex limits easy to solve, especially when the terms converge rapidly due to exponential functions. In our discussion, substituting \(x = \frac{2}{n+1}\) and applying this approximation as \(n\) grows large results in \( e^{-n \cdot \frac{2}{n+1}} \). As \(n\) approaches infinity, this further simplifies to \(e^{-2}\).
Using this approximation makes complex limits easy to solve, especially when the terms converge rapidly due to exponential functions. In our discussion, substituting \(x = \frac{2}{n+1}\) and applying this approximation as \(n\) grows large results in \( e^{-n \cdot \frac{2}{n+1}} \). As \(n\) approaches infinity, this further simplifies to \(e^{-2}\).
- This technique is useful for evaluating limits involving repeated multiplication of similar terms.
- Exponential approximations provide smooth approaches to convergence, often resulting in strikingly simple expressions.
Limit Evaluation Techniques
Effective evaluation of limits often requires employing a mixture of analytical strategies. Techniques such as approximation, transformations, and the recognition of known patterns are pivotal in simplifying expressions that might initially appear daunting. Here’s a summary of tactics:
- Simplification: Reduce the complexity of the expression; this might involve factoring or basic algebraic manipulation, such as recognizing that \(\frac{n-1}{n+1} \approx 1 - \frac{2}{n+1}\) for large \(n\).
- Approximation: Use limit approximations—such as treating small perturbations with exponential or logarithmic approximations—to streamline calculations as we did with exponential limits.
- Transformation: Utilize transformations to change the form of the limit and analyze it from alternate perspectives that might be easier, as seen when converting behavior at infinity to behavior near zero.