Problem 32 - University Calculus: Early Transcendentals

Question

Use I'Hópital's rule to find the limits. $$\lim _{x \rightarrow \infty} \frac{\log _{2} x}{\log _{3}(x+3)}$$

Step-by-Step Solution

Answer

The limit is \(\frac{\ln 3}{\ln 2}\).

Step 1: Verify Indeterminate Form

Identify that both the numerator and denominator of the expression approach infinity as \(x\) approaches infinity. This forms an indeterminate of type \(\frac{\infty}{\infty}\), making it suitable for L'Hópital's rule.

Step 2: Take Derivatives

Find the derivative of the numerator \(\log_{2} x\), which is \(\frac{1}{x \ln 2}\). Similarly, find the derivative of the denominator \(\log_{3}(x+3)\), which is \(\frac{1}{(x+3) \ln 3}\).

Step 3: Apply L'Hópital's Rule

Using L'Hópital's Rule, evaluate the limit: \[\lim _{x o \infty} \frac{\log_{2} x}{\log_{3}(x+3)} = \lim _{x o \infty} \frac{\frac{1}{x \ln 2}}{\frac{1}{(x+3) \ln 3}}\] which simplifies to \[\lim _{x o \infty} \frac{(x+3) \ln 3}{x \ln 2}.\]

Step 4: Simplify and Evaluate Limit

Simplify the limit:\[\lim _{x o \infty} \frac{(x+3) \ln 3}{x \ln 2} = \lim _{x o \infty} \frac{x \ln 3 + 3 \ln 3}{x \ln 2}.\]This can be further simplified to\[\lim _{x o \infty} \left(\frac{\ln 3}{\ln 2} + \frac{3 \ln 3}{x \ln 2}\right).\]As \(x\) approaches infinity, the second term \(\frac{3 \ln 3}{x \ln 2}\) approaches 0, leading to\(\frac{\ln 3}{\ln 2}.\)

Key Concepts

Indeterminate Forms

In calculus, an indeterminate form is a mathematical expression that does not readily point to a clear limit or value. One common type is \(\frac{\infty}{\infty}\). This typically occurs in limits where both the numerator and denominator approach infinity. When faced with this form, we cannot straightforwardly compute the limit because it is unclear how the expressions grow relative to each other.
To overcome this, we often employ methods such as L'Hôpital's Rule, which allows us to use derivatives to find the limit when faced with indeterminate forms. This often simplifies the problem, making it possible to determine the behavior of the function as the variable approaches a certain value.

• Types of indeterminate forms: \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \cdot \infty\), \(\infty - \infty\), \(1^{\infty}\), \(0^{0}\), \(\infty^{0}\).
• These forms appear frequently in the evaluation of limits.
• L'Hôpital’s Rule helps manage many of these situations.

Derivatives

Derivatives measure the rate of change of a function with respect to a variable. They are foundational in calculus and essential in applying L'Hôpital's Rule. In our exercise, we take the derivatives of logarithmic functions.
For example, the derivative of \(\log_{2} x\) is \(\frac{1}{x \ln 2}\), obtained using the chain rule and the fact that \(\frac{d}{dx}\ln x = \frac{1}{x}\). Similarly, for \(\log_{3}(x+3)\), the derivative is \(\frac{1}{(x+3) \ln 3}\).
Calculating these derivatives:

• The derivative helps to transform the original indeterminate form \(\frac{\log_{2} x}{\log_{3}(x+3)}\) into a simpler expression \(\frac{\frac{1}{x \ln 2}}{\frac{1}{(x+3) \ln 3}}\).
• This simplification is key to finding the limit using L'Hôpital's Rule.
• Mastering derivatives of logarithmic functions expands your problem-solving toolkit in calculus.

Infinity Limits

Infinity limits explore the behavior of functions as the variable grows infinitely large. Calculating limits that lead to infinity often results in indeterminate forms, especially \(\frac{\infty}{\infty}\).
In our exercise, both the numerator and denominator approach infinity as \(x\) increases. This is common when dealing with logarithmically based functions because their values increase beyond bounds as \(x\) becomes large.
Key insights about infinity limits:

• These types of limits ask how a function behaves or grows as the input heads towards infinity.
• They help describe asymptotic behavior and end behavior of functions.
• The solution of the exercise demonstrates how we use derivatives to transform and evaluate these limits effectively with L'Hôpital's Rule.

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions. They are essential in calculus, particularly in understanding growth rates, decay, and solving equations involving exponential growth.
In our problem, we deal with the logarithms base 2 and base 3. Logarithms allow us to tackle multiplicative relationships by converting them into additive ones, which are easier to manipulate algebraically.
Features of logarithmic functions:

• They grow slowly compared to polynomials and exponentials.
• Their domains are positive real numbers, but they can approach negative infinity as the input gets closer to zero.
• The derivative of a logarithmic function \(\log_b x\) is \(\frac{1}{x \ln b}\), where \(b\) is the base.
• In our solution, applying derivatives to these functions allows us to transform and simplify preliminary forms into manageable operations.