Problem 31 - Thomas Calculus in SI Units

Question

Use l'Hôpital's rule to find the limits. $$\lim _{x \rightarrow \infty} \frac{\ln (x+1)}{\log _{2} x}$$

Step-by-Step Solution

Answer

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

Step 1: Verify Indeterminate Form

First, we need to check if the limit is in an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). As \( x \to \infty \), \( \ln(x+1) \to \infty \) and \( \log_{2}(x) \to \infty \), which means we have an indeterminate form \( \frac{\infty}{\infty} \).

Step 2: Apply l'Hôpital's Rule

Since we have an indeterminate form \( \frac{\infty}{\infty} \), we can apply l'Hôpital's Rule. Take the derivatives of both the numerator and the denominator:- The derivative of \( \ln(x+1) \) is \( \frac{1}{x+1} \).- The derivative of \( \log_{2}(x) \) is \( \frac{1}{x \ln(2)} \).So, our limit becomes:\[\lim_{x \to \infty} \frac{\frac{1}{x+1}}{\frac{1}{x \ln(2)}}\]

Step 3: Simplify the Expression

Simplify \( \frac{\frac{1}{x+1}}{\frac{1}{x \ln(2)}} \):\[\lim_{x \to \infty} \frac{x \ln(2)}{x+1}\]This simplifies further to:\[\lim_{x \to \infty} \frac{x \ln(2)}{x+1} = \lim_{x \to \infty} \frac{\ln(2)}{1 + \frac{1}{x}}\]

Step 4: Evaluate the Limit

As \( x \to \infty \), \( \frac{1}{x} \to 0 \), so the expression simplifies to:\[\lim_{x \to \infty} \frac{\ln(2)}{1 + \frac{1}{x}} = \ln(2)\]Thus, the limit is \( \ln(2) \).

Key Concepts

Indeterminate Forms

When working with limits, sometimes you encounter forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These are called indeterminate forms.
Indeterminate forms are tricky because they don't clearly convey what the limit actually is. Luckily, mathematicians have tools to deal with them, like l'Hôpital's rule.
In this exercise, we first checked if the expression \( \frac{\ln(x+1)}{\log_{2}(x)} \) leads to an indeterminate form as \( x \) approaches infinity.

• As \( x \to \infty \), \( \ln(x+1) \) heads to infinity.
• Similarly, \( \log_{2}(x) \) also goes to infinity.

Because both parts of the fraction go to infinity, we have the \( \frac{\infty}{\infty} \) form.
You can simplify expressions like this with l'Hôpital's rule. This rule allows us to find limits when faced with these indeterminate situations.

Derivatives

Derivatives are a key tool in calculus that measure how a function changes as its input changes.
When we apply l'Hôpital's rule, derivatives are crucial because we replace the original functions in the limit with their derivatives.
In this problem, we take derivatives of both the numerator and denominator:

• \( \ln(x+1) \) becomes \( \frac{1}{x+1} \);
• \( \log_{2}(x) \) translates to \( \frac{1}{x \ln(2)} \).

This gives a new fraction \( \frac{\frac{1}{x+1}}{\frac{1}{x \ln(2)}} \).
Using derivatives helps us to transform the problem from an indeterminate form to one that we can solve more easily.

Limits

Limits help us understand the behavior of functions as they get very close to a certain point.
In this exercise, we're interested in what happens as \( x \) approaches infinity.
After applying l'Hôpital's rule and finding the derivatives, we derive a new expression:

• \( \lim_{x \to \infty} \frac{x \ln(2)}{x+1} \).

This expression simplifies further to:

• \( \lim_{x \to \infty} \frac{\ln(2)}{1 + \frac{1}{x}} \).

As \( x \to \infty \), \( \frac{1}{x} \to 0 \), simplifying the expression to just \( \ln(2) \).
Understanding limits is key to solving these types of problems because they offer insights into how expressions behave at extreme values.

Natural Logarithm

The natural logarithm, \( \ln(x) \), is a logarithm with the base \( e \), where \( e \approx 2.71828 \).
It is widely used in mathematics because of its unique properties, especially in calculus.
In this exercise, we have \( \ln(x+1) \) in the numerator, which plays a central role in forming the indeterminate form.
Here are a few key properties of \( \ln(x) \) that help in calculus:

• The derivative \( \frac{d}{dx}[\ln(x)] = \frac{1}{x} \), aiding us when used in l'Hôpital's rule.
• As \( x \to \infty \), \( \ln(x) \to \infty \), a key factor in finding that we have an indeterminate form.

Natural logarithms help simplify complex exponential growth expressions, which is why they appear frequently in mathematical limits and calculus.