Problem 57 - Thomas Calculus in SI Units

Question

Find the limits $$\lim _{x \rightarrow \infty}(1+2 x)^{1 /(2 \ln x)}$$

Step-by-Step Solution

Answer

The limit is \( \sqrt{e} \).

Step 1: Consider the function for substitution

The given limit problem is \( \lim_{x \to \infty} (1 + 2x)^{1/(2 \ln x)} \). Evaluate the term inside the limit which is in the form \((1 + f(x))^g(x)\) where \(f(x) = 2x\) and \(g(x) = \frac{1}{2\ln x}\). To simplify this, substitute \( y = 1 + 2x \) and \( x \to \infty \) implies \( y \to \infty \). The expression becomes \( y^{1/(2 \ln(\frac{y-1}{2}))} \).

Step 2: Simplify the expression

The expression simplifies to \( y^{\frac{1}{2 (\ln y - \ln 2)}} \). For large \( y \), \( \ln y - \ln 2 \approx \ln y \). Thus, the expression is approximated by \( y^{\frac{1}{2 \ln y}} \). This can be rewritten using properties of exponents as \( y^{1/(2 \ln y)} = e^{\ln y/(2 \ln y)} \).

Step 3: Evaluate the exponent

Simplify \( \frac{\ln y}{2 \ln y} \) to \( \frac{1}{2} \). So, the expression becomes \( e^{1/2} \).

Step 4: Conclude the limit

Since the expression simplifies to the constant \( e^{1/2} \), which is independent of \( x \), as \( y \to \infty \) (or equivalently \( x \to \infty \)), the limit approaches \( e^{1/2} \). Therefore, the result of the limit evaluation is \( \sqrt{e} \).

Key Concepts

Limit Evaluation

Limit evaluation in calculus is a fundamental concept that helps us to understand the behavior of functions as they approach a certain point or infinity. In this exercise, we are tasked with finding the limit of the function:

• \( \lim \_{x \rightarrow \infty}(1+2x)^{1 /(2 \ln x)} \)

To evaluate this limit, we first notice that the function inside the limit is in the form \((1 + f(x))^g(x)\), where \(f(x) = 2x\) and \(g(x) = \frac{1}{2\ln x}\). The goal is to simplify this expression to a form where we can easily calculate the limit.
We substitute \( y = 1 + 2x \), which recasts the limit in terms of \( y \) as \( y^{1/(2 \ln(y-1/2))} \). As \( x \rightarrow \infty \), \( y \rightarrow \infty \), allowing us to simplify the logarithmic term as \( \ln(y - 1/2) \approx \ln y \) for large \( y \).
Understanding these simplifications and substitutions helps us to reduce complex expressions into more manageable forms, ultimately allowing us to find the limit with greater ease and clarity. By continuing the simplification process, we reach the result of the limit as \( \sqrt{e} \).

Exponential Functions

Exponential functions are essential in this exercise as they play a key role in the transformation and simplification process. An exponential function is a mathematical expression where a constant base is raised to a variable exponent.
In our problem, after substituting and simplifying, the expression \( y^{1/(2 \ln y)} \) is transformed to an exponential form using properties of logarithms and exponents:

• \( e^{\ln y/(2 \ln y)} \)

This step of transforming the power into an exponent is crucial because it facilitates the evaluation of the limit by reducing the complex exponent into a simple constant.
The power simplification means finding \( \frac{\ln y}{2 \ln y} \), which straightforwardly reduces to \( \frac{1}{2} \). This simplification is where the logarithm rules become handy, especially the property that \( a^{b/c} = e^{(b/c)\ln a} \).
Hence, the power of \( e \) becomes \( e^{1/2} \), which is equivalent to \( \sqrt{e} \), showcasing the power of exponential functions in transforming and evaluating limits.

Asymptotic Behavior

Asymptotic behavior describes how a function behaves as its argument either grows very large or very small. In this problem, the function \((1 + 2x)^{1/(2\ln x)}\) is analyzed as \(x\) approaches infinity. Asymptotic analysis helps identify dominant terms and aid in understanding long-term behavior.
Substituting \(y = 1 + 2x\) gives the transformed expression's asymptotic behavior. As \(x \rightarrow \infty\), so does \(y\). This makes \( \ln (y-1/2) \approx \ln y \) valid for large values, allowing reduction in terms. Such behavior allows simplification using approximations, as seen where \(\ln (y-1/2)\) nearly matches \(\ln y\), smoothing the path to completing limit evaluations.
We see that approaching infinity reduces complexities to constant results. This method demonstrates how convoluted expressions can be managed using asymptotic analysis. Ultimately, these approaches provide a detailed understanding of how different portions of a function interact and maintain simplify evaluation as seen through our conclusion \(\sqrt{e}\).