Problem 20 - Thomas Calculus in SI Units

Question

Use l'Hôpital's rule to find the limits. $$\lim _{x \rightarrow 1} \frac{x-1}{\ln x-\sin \pi x}$$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{1 + \pi} \).

Step 1: Verify Indeterminate Form

First, substitute \( x = 1 \) into the given function: \( \frac{x-1}{\ln x - \sin \pi x} \). This gives \( \frac{1-1}{\ln 1 - \sin \pi \cdot 1} \), which simplifies to \( \frac{0}{0} \). Since we have a \( \frac{0}{0} \) indeterminate form, we can apply l'Hôpital's Rule.

Step 2: Differentiate the Numerator

Differentiate the numerator \( x-1 \). The derivative is \( 1 \).

Step 3: Differentiate the Denominator

Differentiate the denominator \( \ln x - \sin \pi x \). The derivative is \( \frac{1}{x} - \pi \cos \pi x \).

Step 4: Apply l'Hôpital's Rule

Apply l'Hôpital's Rule by taking the limit of the derivatives:\[\lim_{x \to 1} \frac{1}{\frac{1}{x} - \pi \cos \pi x}\]. Substitute \( x = 1 \) into the derivatives to get:\[\frac{1}{\frac{1}{1} - \pi \cos \pi \cdot 1} = \frac{1}{1 - (-\pi)} = \frac{1}{1 + \pi}\].

Step 5: Simplify the Result

Simplify the expression \( \frac{1}{1 + \pi} \). This is already simplified since it is a numerical value.

Key Concepts

Indeterminate Form

When calculating limits in calculus, you might encounter expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These are known as indeterminate forms. They don't have a straightforward value, making direct substitution of the limit difficult to evaluate. Instead of giving up, these forms signal us to find another approach to solve the problem. L'Hôpital's rule is specifically designed to handle such situations, allowing us to apply derivatives to find the limit without losing important data in the simplification process. In the given problem, substituting \( x = 1 \) right away results in \( \frac{0}{0} \), which clearly tells us to transform the equation using l'Hôpital's rule.

Derivatives

Derivatives are a fundamental tool in calculus for understanding how functions change. They provide a way to calculate the slope or rate of change of a function at any given point. This quality is especially useful in l'Hôpital's rule, where derivatives are used to work around indeterminate forms.When confronted with \( \frac{0}{0} \), we perform differentiation on both the numerator and the denominator. - In the exercise, the numerator \( x-1 \) is simple: its derivative is 1. - For the denominator \( \ln x - \sin \pi x \), the derivative becomes \( \frac{1}{x} - \pi \cos \pi x \). These new functions from differentiation replace the original ones in the limit calculation. Once this is done, we can re-evaluate the limit to find a more defined result.

Limits

Limits help us understand the behavior of a function as it approaches a specific value. In calculus, finding limits is crucial because it gives insight into the function's continuity and stability at points that might not be straightforward by normal arithmetic.In our problem, we need to find the limit of \( \frac{x-1}{\ln x - \sin \pi x} \) as \( x \) approaches 1. Using l'Hôpital's rule and derivatives, we reformulate the expression to \( \lim_{x \to 1} \frac{1}{\frac{1}{x} - \pi \cos \pi x} \). Now, substituting \( x = 1 \) simplifies the computation and allows us to evaluate the new form directly without facing an indeterminate result.

Calculus Problem Solving

Calculus is a branch of mathematics that deals with change and is reinforced by problem-solving techniques like limits, derivatives, and integrals. Approaching problems systematically using these techniques provides clarity in cases of indeterminate expressions.Problem solving in calculus often involves:- **Identifying potential formulas like indeterminate forms:** Noticing \( \frac{0}{0} \) directs us toward l'Hôpital's rule.- **Using derivatives for simplification:** Differentiating helps change troublesome terms into simpler ones.- **Substitution and simplification:** Post-differentiation, substituting values back helps in arriving at a defined numerical result.In our exercise, this sequence helps simplify and solve \( \lim_{x \to 1} \frac{x-1}{\ln x - \sin \pi x} \), resulting in the final answer of \( \frac{1}{1 + \pi} \). Employing this method accurately demonstrates calculus' power in problem resolving scenarios.