Question

Find each limit. Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$ \lim _{x \rightarrow \infty}[\ln (x+1)-\ln (x-1)] $$

Short answer

The limit is 0.

Step-by-step solution

Evaluate the Original Expression

Consider the expression \( \lim _{x \rightarrow \infty}[\ln (x+1)-\ln (x-1)] \). Initially, let's simplify the expression using the properties of logarithms. The difference of logarithms can be rewritten as: \[ \ln(x+1) - \ln(x-1) = \ln\left(\frac{x+1}{x-1}\right) \].

Simplify the Expression

Now simplify the fraction inside the logarithm: \( \frac{x+1}{x-1} = \frac{x+1}{x-1} \). For very large \( x \), rewrite each term inside the fraction as: \[ \frac{x(1 + \frac{1}{x})}{x(1 - \frac{1}{x})} = \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}} \].

Determine the Indeterminate Form

As \( x \rightarrow \infty \), both the numerator \((1 + \frac{1}{x})\) and the denominator \((1 - \frac{1}{x})\) approach 1. Therefore, \( \ln\left(\frac{1 + \frac{1}{x}}{1 - \frac{1}{x}}\right) \) becomes an indeterminate form \( \frac{0}{0} \) when considering logarithmic expansion. Check indeterminate form to apply l'Hôpital's Rule.

Apply l'Hôpital's Rule

Since we have identified the indeterminate form, apply l'Hôpital's Rule, which involves differentiating the numerator and the denominator. Differentiate \( x+1 \) and \( x-1 \) with respect to \( x \):\[ \lim_{x \to \infty} \frac{\frac{d}{dx}[x+1]}{\frac{d}{dx}[x-1]} \].The derivatives are:- Numerator: derivative of \( x+1 \) is 1- Denominator: derivative of \( x-1 \) is also 1.

Evaluate the Limit After Differentiation

Now the limit becomes:\[ \lim_{x \to \infty} \frac{1}{1} = 1. \]Hence, the limit of the original logarithmic expression is \( 0 \) because the logarithm of 1 is 0.

Key concepts

l'Hôpital's Rule

l'Hôpital's Rule is a powerful tool in calculus used for finding limits of indeterminate forms, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). It simplifies the process by transforming the indeterminate form into a more intuitive expression. Here's how it works:

• Check that the limit results in an indeterminate form. This step is crucial before applying the rule.
• Differentiating both the numerator and the denominator separately. This is usually achieved with basic differentiation rules.
• Take the limit of the new fraction obtained from the derivatives.

For instance, in our example,
\( \lim_{x \to \infty} \ln \left( \frac{x+1}{x-1} \right) \)
appears first as an indeterminate form \(\frac{0}{0}\). After verifying its form, we apply l'Hôpital's Rule by differentiating:

• Numerator becomes: derivative of \(x+1\) is \(1\)
• Denominator becomes: derivative of \(x-1\) is \(1\)

Finally, evaluate the simple limit of \(\frac{1}{1}\), which results in \(1\). Remember, the logarithm of \(1\) is \(0\), leading to the final solution.

Logarithmic Functions

Logarithmic functions allow us to explore areas involving exponential growth or decay. They are the inverse of exponential functions. Here's what you need to remember about them:

• Logarithms help simplify multiplication into addition and division into subtraction.
• Formally, if \(a^b = c\), then the logarithm of \(c\) with base \(a\) is \(b\), which we write as \(\log_a(c) = b\).
• The natural logarithm \(\ln(x)\) uses base \(e\), where \(e\) is Euler’s number, approximately \(2.718\).

In the expression we're examining, we have
\(\ln(x+1) - \ln(x-1)\),
which uses the property of logarithms that allows us to convert it to one logarithm:
\(\ln\left( \frac{x+1}{x-1} \right)\).
This conversion simplifies the problem to actively use calculus techniques.

Indeterminate Forms

Indeterminate forms arise in calculus when an expression does not lead to a definitive limit and instead appears contradictory or ambiguous. The most common types are:

• \(\frac{0}{0}\)
• \(\frac{\infty}{\infty}\)
• \(0 \times \infty\)

These indeterminate forms signal the potential to apply l'Hôpital's Rule. Let's consider why our example turns into an indeterminate form. By rewriting
\( \ln\left( \frac{x+1}{x-1} \right) \)
and simplifying, as \(x\) approaches infinity, both parts inside our fraction approach one,
turning our expression into \(\ln(1)\), which doesn't directly offer the limit.
This behavior shows the path for using advanced calculus techniques, such as l'Hôpital's Rule, to resolve the ambiguity and find the precise limit.