Problem 37 - University Calculus: Early Transcendentals

Question

Use I'Hópital's rule to find the limits. $4\lim _{x \rightarrow \infty}(\ln 2 x-\ln (x+1))$$

Step-by-Step Solution

Answer

The limit is \( 4\ln(2) \).

Step 1: Recognize the Indeterminate Form

We are given the limit \( 4 \lim_{x \to \infty} (\ln 2x - \ln(x+1)) \). As \( x \to \infty \), both \( \ln 2x \) and \( \ln(x+1) \) approach infinity, hence the expression takes the form \( \infty - \infty \). To apply L'Hôpital's Rule, we need to convert it to a suitable indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).

Step 2: Combine the Logarithms

Use the properties of logarithms to combine the expression into a single logarithm: \( \ln 2x - \ln(x+1) = \ln\left(\frac{2x}{x+1}\right) \). This simplifies our expression and allows us to reconsider the form of the limit as \( x \to \infty \).

Step 3: Set Up the Limit

Rewrite the limit: \( 4 \lim_{x \to \infty} \ln\left(\frac{2x}{x+1}\right) \). We need to reconsider the form \( \frac{2x}{x+1} \) as \( x \to \infty \).

Step 4: Simplify the Inside of the Logarithm

Inside the logarithm, as \( x \to \infty \), \( \frac{2x}{x+1} \approx \frac{2x}{x} = 2 \). So, the argument of the logarithm approaches 2 as \( x \to \infty \).

Step 5: Evaluate the Limit

Evaluate \( \lim_{x \to \infty} \ln\left( \frac{2x}{x+1} \right) \), which approaches \( \ln(2) \) because \( \frac{2x}{x+1} \to 2 \).

Step 6: Multiply by the Coefficient

Finally, multiply the result by 4: \( 4 \cdot \ln(2) \).

Step 7: Conclusion

The original limit is \( 4 \cdot \ln(2) \). Therefore, the limit simplifies to \( 4\ln(2) \).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful method in calculus used to solve indeterminate forms that arise while finding limits. In cases where you have forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), this rule is very helpful. Here's a simple way to think of it:

• If a limit leads to an indeterminate form, take the derivative of the numerator and the derivative of the denominator separately.
• Then, calculate the limit of the resulting expression.

Always remember, L'Hôpital's Rule can only be applied when the initial limit is an indeterminate form of \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Therefore, your first step should be identifying the form of the limit. In our exercise, the original expression was turned into one of these indeterminate forms using logarithm properties, allowing the application of L'Hôpital's Rule. Habits like rewriting expressions and using differentiation make it possible to tackle these problems efficiently.

Indeterminate Forms

In calculus, an "indeterminate form" refers to expressions where direct substitution to find a limit doesn’t yield a specific value. Common indeterminate forms include:

• \( \frac{0}{0} \)
• \( \frac{\infty}{\infty} \)
• \( \infty - \infty \)
• \( 0^0 \)

When faced with such forms, a limit can be tricky because it doesn't point to an immediate conclusion.
In the given exercise, the expression \( \ln 2x - \ln(x+1) \) approached \( \infty - \infty \), an indeterminate form. Our primary task was to transform this into an appropriate form, suitable for L'Hôpital's Rule, like \( \frac{\infty}{\infty} \). This was achieved by using the property of logarithms to combine the terms, drastically simplifying the calculation.

Logarithm Properties

Logarithm properties can greatly simplify complex expressions, making them manageable for calculus students. One key property is the difference of logarithms:

• \( \ln a - \ln b = \ln\left(\frac{a}{b}\right) \).

This property was used in our exercise to simplify a subtraction of logarithms into a single logarithm of a quotient.
Recognizing these properties can transform a challenging problem into something straightforward. When calculating limits involving logarithms, first see if you can apply these properties to make the problem cleaner. For the exercise, we successfully reduced \( \ln 2x - \ln(x+1) \) into \( \ln\left(\frac{2x}{x+1}\right) \), a simpler expression that readily allowed the use of further calculus techniques, such as L'Hôpital's Rule, to evaluate the limit.