Problem 60 - Calculus Early Transcendentals

Question

Evaluate the following limits or explain why they do not exist. Check your results by graphing. $$\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{\ln x}$$

Step-by-Step Solution

Answer

Answer: The limit as x approaches infinity of the function (1 + 1/x)^(ln x) is 1.

Step 1: Rewrite the limit using properties of logarithms and exponential functions

The given expression can be rewritten as: $$\lim _{x \rightarrow \infty} e^{\ln\left({\left(1+\frac{1}{x}\right)^{\ln x}}\right)}$$

Step 2: Simplify the expression inside the limit

Apply the logarithm properties to simplify the exponent part: $$\lim _{x \rightarrow \infty} e^{\ln x \cdot \ln\left(1+\frac{1}{x}\right)}$$ Let \(y=\ln x\). As \(x \rightarrow \infty\), then \(y = \ln x \rightarrow \infty\) So, the given limit becomes: $$\lim _{y \rightarrow \infty} e^{y \cdot \ln\left(1+\frac{1}{e^y}\right)}$$

Step 3: Apply l'Hôpital's Rule

Rewrite the exponent part as a fraction and apply l'Hôpital's Rule: $$\lim _{y \rightarrow \infty} e^{\frac{\ln\left(1+\frac{1}{e^y}\right)}{\frac{1}{y}}}=e^{\lim _{y \rightarrow \infty}\frac{\ln\left(1+\frac{1}{e^y}\right)}{\frac{1}{y}}}$$ Differentiate both numerator and denominator with respect to \(y\): $$ e^{\lim _{y \rightarrow \infty}\frac{-\frac{1}{e^y}}{1+\frac{1}{e^y}}\cdot\frac{-e^y}{y^2}}= e^{\lim _{y \rightarrow \infty}\frac{1}{(1+\frac{1}{e^y})\cdot y}}$$

Step 4: Evaluate the limit of the simplified expression

As \(y \rightarrow \infty\), the limit becomes: $$ e^{\lim _{y \rightarrow \infty}\frac{1}{(1+0)\cdot y}}=e^{\lim _{y \rightarrow \infty}\frac{1}{y}}$$ Since \(\lim _{y \rightarrow \infty}\frac{1}{y} = 0\): $$ e^{\lim _{y \rightarrow \infty}\frac{1}{y}}= e^0$$ Finally, we get: $$ e^0 = 1$$ Thus, the limit is: $$\lim_{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{\ln x} = 1$$

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a vital tool in calculus, especially when faced with limits that result in indeterminate forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). The essence of this rule is to differentiate the numerator and the denominator separately. The limit of the resulting fraction can then sometimes be easier to evaluate.
In our given exercise, we applied L'Hôpital's Rule to the limit:

• \(\lim _{y \rightarrow \infty}\frac{\ln\left(1+\frac{1}{e^y}\right)}{\frac{1}{y}}\)

By differentiating the numerator \(\ln(1+\frac{1}{e^y})\) and denominator \(\frac{1}{y}\), it becomes simpler to evaluate. This approach transforms complex limits into solvable problems, essential for tackling advanced calculus challenges.
Always ensure the application conditions are met before using L'Hôpital's Rule, and remember there are alternative methods to solve limits if L'Hôpital's Rule does not simplify the expression adequately.

Exponential Functions

Exponential functions are foundational to calculus, characterized by their constant rate of growth or decay. The function \(e^x\) is a prime example, representing continuous growth and playing a crucial role in compound interest and natural processes.
In our problem, we used the property that

• \(\left(1+\frac{1}{x}\right)^{\ln x}\)

can be rewritten using exponentiation:\(\lim _{x \rightarrow \infty} e^{\ln\left(\left(1+\frac{1}{x}\right)^{\ln x}\right)}\). By transforming expressions into their exponential form, they often become easier to work with, especially when applying logarithmic properties and L'Hôpital's Rule.
Understanding the behavior of these functions at their limits is key, as it allows for the interpretation of real-world phenomena such as population growth and radioactive decay.

Logarithmic Properties

Logarithms are often useful in simplifying complex expressions, especially when they're presented in an exponential form. In calculus, they help to manage expressions with powers and roots.
In our exercise, the expression:

• \(\ln\left(\left(1+\frac{1}{x}\right)^{\ln x}\right)\)

was greatly simplified by using the logarithmic property \(\ln(a^b) = b\ln(a)\). This turned the inner part of the original limit into a manageable form: \(\ln x \cdot \ln\left(1+\frac{1}{x}\right)\).
Logarithmic transformations often reveal patterns and simplify differentiations and integrations, hence they're a power tool for solving calculus problems smoothly.
It's beneficial to be comfortable with key logarithmic identities. These identities aid in identifying and applying the most efficient technique for evaluating intricate limits.