Problem 38 - University Calculus: Early Transcendentals

Question

Use I'Hópital's rule to find the limits. $$\lim _{x \rightarrow 0^{-}}(\ln x-\ln \sin x)$$

Step-by-Step Solution

Answer

The limit is 0 as x approaches 0 from the negative side.

Step 1: Identify the Form

We have the limit \( \lim_{x \to 0^{-}} (\ln x - \ln \sin x) \). This can be rewritten as \( \ln \left(\frac{x}{\sin x}\right) \). The expression \( \frac{x}{\sin x} \) approaches \( \infty \) as \( x \rightarrow 0^{-} \), so the expression becomes \( -\infty \). However, to correctly apply L'Hôpital's Rule, let's verify the form by expressing it as a quotient: \[ \frac{\ln x - \ln \sin x}{1} \]. Substitute to see the indeterminate form \( \frac{-\infty}{1} = -\infty \), which is not directly suitable for L'Hôpital's Rule.

Step 2: Rewrite the Expression

To prepare for L'Hôpital's Rule, consider rewriting \( \ln \left(\frac{x}{\sin x}\right) \) as a single expression that involves a quotient. We can set it up as a division: \( \lim_{x \to 0^{-}} \frac{(\ln x - \ln \sin x)}{1} \). However, as seen previously, this needs further adjustment as it was in determinate form and does not suit L'Hôpital's Rule directly.

Step 3: Apply L'Hôpital's Rule

Notice there was a mistake in identifying the form; actually, analyzing separately without direct L'Hôpital's seems more consistent. Return and check \( \ln x - \ln \sin x = \ln \left(\frac{x}{\sin x}\right) \) considering direct behavior without direct L'Hôpital straigh away as formal rewriting was problematic as \( x \to 0^- \) concerns separable check.

Step 4: Express Limit Using Taylor Expansions to Simplify

As \( x \rightarrow 0^{-} \), both \( \ln x \) and \( \ln \sin x \) move towards \( -\infty \), use expansion: Recall \( \sin x \approx x - \frac{x^3}{6} \) for small \( x \), so \( \frac{x}{\sin x} \approx 1 + \frac{x^2}{6} \). Hence, \( \ln \left(\frac{x}{\sin x}\right) \approx \ln (1 + \frac{x^2}{6}) \sim \frac{x^{2}}{6} \). Since \( x \rightarrow 0^{-} \), substitute back to find:\[ \lim_{x \to 0^-} \ln \left(1 + \frac{x^2}{6} \right) \approx 0 \].

Step 5: Conclusion

The limit \( \lim_{x \to 0^-} \ln \left(\frac{x}{\sin x}\right) \) as derived eturns 0 since using more cautious Taylor rather than rote-quite L'Hôpital directly; hence no deviation in higher orders shifts directly outcome.

Key Concepts

Limits

The concept of limits is crucial in calculus. It helps us understand the behavior of functions as they approach a specific point. In this exercise, we're looking at the limit as \(x\) approaches \(0^{-}\) for the function \( \ln x - \ln \sin x \). This expression can be rewritten as \( \ln \left(\frac{x}{\sin x}\right) \). Calculating limits like this one involves determining what happens to the function as \(x\) gets very close to the value, but not exactly at it. This tells us the trend or direction in which the function is heading. Understanding limits allows us to tackle problems where direct substitution leads to undefined or indeterminate forms.

Indeterminate Forms

Indeterminate forms arise in limit problems that initially appear undefined. They're expressions like \(\frac{0}{0}\) or \(\frac{-\infty}{\infty}\) that signal more work is needed to find the actual limit. In this problem, rewriting \( \ln x - \ln \sin x \) gives \(\frac{-\infty}{1}\) as \(x\) approaches \(0^{-}\), which suggests an undefined result at first glance. We use indeterminate forms to remind ourselves that we can't always just plug in values to find limits. Instead, these forms direct us to apply advanced techniques, such as rewriting expressions or using L'Hôpital's Rule, to achieve meaningful results.

Taylor Series

Taylor series breaks down complicated functions into infinitely long polynomials to make them easier to handle. By using a Taylor series, we can approximate functions near a point with higher accuracy. For instance, in this exercise, the series expansion for \( \sin x \) is used to simplify the expression \( \frac{x}{\sin x} \). Near \(x = 0\), \(\sin x\) can be approximated as \(x - \frac{x^3}{6}\). This simplifies \( \frac{x}{\sin x} \) to approximately \(1 + \frac{x^2}{6}\), which is then plugged back into the logarithmic function to estimate the limit. This approximation eases the process and helps in revealing the limit as \(0\).

Logarithmic Functions

Logarithmic functions have unique characteristics that are useful in calculus. The natural logarithm, \(\ln x\), is logarithmic to the base \(e\) and is particularly significant due to its properties. In this problem, we are using properties of logarithms, like \(\ln a - \ln b = \ln\left(\frac{a}{b}\right)\), to transform and simplify the original expression. This approach interplays with Taylor series to further simplify the function \( \ln \left( \frac{x}{\sin x} \right) \), allowing us to apply limits more effectively. By understanding these logarithmic relationships, we can break down complex limits into something more manageable and comprehensible.