Problem 32 - Thomas Calculus in SI Units

Question

Use l'Hôpital'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 \(\log_{2} 3\).

Step 1: Recognize the Indeterminate Form

First, we need to identify if the limit presents an indeterminate form suitable for l'Hôpital's rule. As \(x \rightarrow \infty\), both the numerator \(\log_{2} x\) and the denominator \(\log_{3}(x+3)\) approach infinity, thus the limit is of the form \(\frac{\infty}{\infty}\). This allows us to apply l'Hôpital's rule.

Step 2: Apply l'Hôpital's Rule

According to l'Hôpital's rule, taking the derivative of the numerator and the derivative of the denominator should resolve the indeterminate form.The derivative of \(\log_{2} x\) is \(\frac{1}{x \ln 2}\), and the derivative of \(\log_{3}(x+3)\) is \(\frac{1}{(x+3) \ln 3}\).

Step 3: Compute the New Limit

Now, substitute the derivatives back into the limit:\[\lim_{x \rightarrow \infty} \frac{\frac{1}{x \ln 2}}{\frac{1}{(x+3) \ln 3}} = \lim_{x \rightarrow \infty} \frac{(x+3) \ln 3}{x \ln 2}.\]

Step 4: Simplify the Expression

Expand and simplify the fraction:\[\lim_{x \rightarrow \infty} \frac{x \ln 3 + 3 \ln 3}{x \ln 2} = \lim_{x \rightarrow \infty} \left(\frac{x \ln 3}{x \ln 2} + \frac{3 \ln 3}{x \ln 2}\right).\]This simplifies to:\[\lim_{x \rightarrow \infty} \frac{\ln 3}{\ln 2} + \lim_{x \rightarrow \infty} \frac{3 \ln 3}{x \ln 2}.\]

Step 5: Evaluate the Simplified Limits

The term \(\frac{3 \ln 3}{x \ln 2}\) approaches 0 as \(x \rightarrow \infty\) since \(x\) is in the denominator. Thus, the overall limit is:\[\frac{\ln 3}{\ln 2} + 0 = \frac{\ln 3}{\ln 2}.\]

Step 6: Express the Final Answer

Since \(\frac{\ln 3}{\ln 2}\) is equivalent to \(\log_{2} 3\) using the change of base formula, the limit is equal to \(\log_{2} 3\).

Key Concepts

Indeterminate Forms

In calculus, "indeterminate forms" are expressions that do not initially have a clear limit or outcome just by substitution. They often occur during limit calculations, especially when both the numerator and denominator approach infinity, zero, or exhibit other behaviors that complicate direct evaluation. Imagine trying to calculate a limit of the form \(\frac{f(x)}{g(x)}\) as \(x\) approaches a certain value. If both \(f(x)\) and \(g(x)\) tend to infinity or zero, we're dealing with an indeterminate form.In the original exercise, as \(x\) approaches infinity, both \(\log_2 x\) and \(\log_3(x+3)\) tend toward infinity, leading to an \(\frac{\infty}{\infty}\) situation. This specific form is one of the classic scenarios where l’Hôpital's Rule becomes useful. Addressing these forms properly allows for precise calculation of limits, bypassing misleading initial evaluations. Understanding these forms helps in applying rules like l'Hôpital's effectively, ensuring mathematical precision.

Calculus Limits

Calculus limits explore the behavior of functions as their inputs approach a certain value, often resulting in insights about continuity, rates of change, and boundless behaviors. Limits are fundamental in understanding seamless change and are the building blocks of derivatives and integrals. Specifically, in this scenario where \(x\) tends to infinity, we analyze how the function's behavior evolves as \(x\) grows indefinitely.The example in the exercise is a great illustration:

• The numerator \(\log_2 x\) and the denominator \(\log_3(x+3)\) both become larger indefinitely as \(x\) increases, which informs us about the overall tendency of the expression.
• This ability to explore large-scale behaviors offers meaningful insights, allowing mathematicians to predict outcomes across different domains, from pure mathematics to applied sciences.

Mastering these concepts enables students to solve various calculus problems efficiently, an essential skill in both theoretical and applied mathematics.

Derivative Calculations

Derivatives are central to calculus, providing a way to understand the rate at which one quantity changes relative to another. They are defined as the limit of the average rate of change as the interval approaches zero. In applying l'Hôpital's Rule, derivatives play a critical role in transforming indeterminate forms into determinate ones that are easier to manage.In the original exercise, we computed derivatives for the logarithmic functions:

• The derivative of \(\log_{2} x\) is \(\frac{1}{x \ln 2}\), found using the property that \(\frac{d}{dx} [\ln_b x] = \frac{1}{x \ln b}\).
• Similarly, the derivative of \(\log_{3} (x+3)\) is \(\frac{1}{(x+3) \ln 3}\).

By substituting these derivatives back into the limit, we simplify the expression, finding new paths to understanding the function's behavior at infinity. This showcases the power of derivatives not only in calculating rates but also in simplifying complex problems such as those presented by indeterminate forms.