Problem 20 - University Calculus: Early Transcendentals

Question

Use I'Hópital'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: Identify the form of the limit

The limit expression is \( \lim_{x \to 1} \frac{x-1}{\ln x - \sin \pi x} \). By substituting \(x = 1\) into the expression, we get \(\frac{1-1}{0 - 0} = \frac{0}{0}\), which is an indeterminate form. This means we can apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule

L'Hôpital's Rule states that for limits of indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we can take the derivative of the numerator and the denominator separately. Differentiating the numerator \(x-1\) gives \(1\). Differentiating the denominator \(\ln x - \sin \pi x\) gives \(\frac{1}{x} - \pi \cos \pi x\).

Step 3: Evaluate the new limit

We now need to find \(\lim_{x \to 1} \frac{1}{\frac{1}{x} - \pi \cos(\pi x)}\). Substitute \(x = 1\) into this expression: \(\frac{1}{1 - \pi \cdot (-1)} = \frac{1}{1 + \pi}\).

Step 4: Simplify and conclude

The new expression is not an indeterminate form and can be evaluated directly as \(\frac{1}{1 + \pi}\). This is the value of the limit.

Key Concepts

Understanding Limits

Limits help us determine the value that a function approaches as the input approaches a certain value. They are essential in calculus and help us make sense of points where the function might not be explicitly defined. In the exercise, we were asked to find \[ \lim _{x \rightarrow 1} \frac{x-1}{\ln x-\sin \pi x} \]which involves determining what happens to the function as \( x \) gets closer and closer to 1.
When we substitute \( x = 1 \), the function evaluates to the indeterminate form \( \frac{0}{0} \). Indeterminate forms are special cases where the limit can't be straightforwardly evaluated using substitution alone. In such cases, tools like L'Hôpital's Rule become invaluable.
Understanding how limits behave and how to manipulate them allows us to work with functions near points where they might not be well-behaved. This opens doors to further calculus concepts like continuity and differentiability.

Indeterminate Forms and L'Hôpital's Rule

Indeterminate forms like \( \frac{0}{0} \), as seen in our exercise, arise when both the numerator and denominator of a fraction approach zero as \( x \to a \). This can be confusing, as it's unclear what the fraction's value should be at that point.
L'Hôpital's Rule is a powerful technique for dealing with these situations. It states that for limits resulting in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can take the derivative of both the numerator and the denominator separately to find the limit. This transformation often removes the indeterminacy.
In the exercise:

• We first identified the \( \frac{0}{0} \) form when substituting \( x = 1 \).
• Then, we applied L'Hôpital's Rule, finding derivatives of both the numerator \( x-1 \) and the denominator \( \ln x - \sin \pi x \).
• This gave us a new expression for which we could find the limit without indeterminate forms.

This process highlights how calculus enables us to evaluate complex limits efficiently.

The Role of Derivatives

Derivatives are fundamental concepts in calculus, providing us insight into how functions change. They are essential tools, especially when dealing with limits and evaluating indeterminate forms.
In applying L'Hôpital's Rule, the derivative allows us to simplify the limit expression. For the exercise, we took the derivative of the numerator \( x-1 \) and the denominator \( \ln x - \sin \pi x \):

• The derivative of \( x-1 \) is 1, since the slope of a linear function \( x \) is constant.
• The derivative of \( \ln x \) is \( \frac{1}{x} \), while the derivative of \(-\sin \pi x \) is \(-\pi \cos \pi x \), using the chain rule.

When we substituted these derivatives back into the limit expression, it offered a straightforward evaluation. This demonstrates how derivatives simplify complex expressions, enabling us to find limits even in initially challenging indeterminate forms.