Problem 35 - Calculus Single Variable

Question

Use the logarithm to reduce the given limit to one that can be handled with l'Hôpital's Rule. \(\lim _{x \rightarrow \infty}\left(\frac{x-5}{x}\right)^{x}\)

Step-by-Step Solution

Answer

The limit is \( e^{-5} \).

Step 1: Identify the Form

Observe that the expression \( \lim _{x ightarrow \infty}\left(\frac{x-5}{x}\right)^{x}\) is in an indeterminate form of \(1^\infty\). This suggests using logarithms to simplify.

Step 2: Transform with the Natural Logarithm

Apply the natural logarithm to the expression by letting \( y = \left(\frac{x-5}{x}\right)^{x} \). Thus, take the natural logarithm: \( \ln y = \ln \left(\left(\frac{x-5}{x}\right)^{x}\right) = x \cdot \ln \left(\frac{x-5}{x}\right) \).

Step 3: Simplify the Logarithmic Expression

Simplify further: \( \ln \left(\frac{x-5}{x}\right) = \ln (x-5) - \ln(x) \), leading to \( x \cdot \left( \ln(x-5) - \ln(x) \right) \).

Step 4: Apply Limit to Logarithmic Expression

Analyze \( \lim_{x \rightarrow \infty} x \cdot \left( \ln(x-5) - \ln(x) \right) \), which can be rewritten as \( \lim_{x \rightarrow \infty} x \left( \ln\left(1 - \frac{5}{x}\right) \right) \).

Step 5: Use Logarithm Approximation

For small values, \( \ln(1 - u) \approx -u \). So, \( \ln\left(1 - \frac{5}{x}\right) \approx -\frac{5}{x} \), giving \( \lim_{x \rightarrow \infty} x \left(-\frac{5}{x}\right) \).

Step 6: Evaluate Limit Using Basic Calculus

The limit simplifies to \( \lim_{x \rightarrow \infty} -5 \), which is simply \(-5\).

Step 7: Interpret the Result

Original limit was transformed to a logarithmic form, resulting in \( \ln y = -5 \). Solve by exponentiating: \( y = e^{-5} \).

Key Concepts

Indeterminate Forms

When evaluating limits, sometimes we encounter expressions that do not have a clear limit. These are called indeterminate forms. One classic example is the form \(1^\infty\). This arises when a function approaches 1 while the exponent tends to infinity, making the behavior unpredictable.
This situation needs careful handling, often using mathematical tools like logarithms or special rules such as l'Hôpital's Rule.
Indeterminate forms require manipulation to determine the actual limit. Techniques often involve algebraic manipulation or applying advanced calculus techniques to handle these forms efficiently.

Natural Logarithms

Natural logarithms are a powerful tool in calculus for simplifying limits, especially those in indeterminate forms. The natural logarithm (\(\ln\)) is the logarithm to the base \(e\), where \(e\) is a fundamental mathematical constant approximately equal to 2.718.
By using natural logarithms, complex expressions can be transformed into more manageable forms. In the problem of \( \lim _{x \rightarrow \infty}\left(\frac{x-5}{x}\right)^{x}\), applying \(\ln\) helps to transition from the power form \(1^\infty\) into a product form that is simpler to analyze mathematically.
Logarithms have the advantage of turning products into sums and powers into products, making them ideal for simplifying tricky limits.

Limit Evaluation

Finding the limit of a function as another variable approaches a certain value is a fundamental task in calculus. Limit evaluation often involves techniques that simplify the expression so it can be solved using simple arithmetic or known rules.
In the exercise, to evaluate \( \lim_{x \rightarrow \infty} x \cdot \ln \left(1 - \frac{5}{x}\right) \), advanced knowledge like l'Hôpital's Rule or approximations of logarithmic expressions for small arguments are frequently used.

• Rewriting tricky expressions into simpler forms can often break down complicated limits into components that are easier to manage.
• Applying known limits, such as \(\ln(1+u) \approx u\) for small \(u\), provides a straightforward path to determining extreme behavior.

Ultimately, the solution involves re-expressing the limit in a form that can be cleanly calculated, showing the beauty of mathematical problem-solving.

Exponential Functions

Exponential functions, with their constant base and variable exponent, are key players in various calculus problems, especially those dealing with growth and decay.
In the final step, reverting back the natural logarithmic transformation involves using the base \(e\) of exponential functions. Given \(\ln y = -5\), we find \(y = e^{-5}\) by exponentiating both sides.

• This operation essentially "undoes" the logarithmic operation, yielding the solution in its original exponential form.
• Exponential growth and decay are universal concepts appearing across mathematical disciplines, highlighting their importance and utility in solving limits.

The elegance of exponential functions lies in their inherent properties, which are techniques that come in handy when resolving such indeterminate form problems.