Question

Use a calculator to graph the function and estimate the value of the limit, then use L'Hôpital's rule to find the limit directly. $$ \lim _{x \rightarrow 1} \frac{x-1}{1-\cos (\pi x)} $$

Short answer

The limit is 0.

Step-by-step solution

Graph the Function

Use a graphing calculator or graphing software to plot the function \( y = \frac{x-1}{1-\cos(\pi x)} \). Observe the behavior of the function as \( x \) approaches 1. This will help provide an estimate for the limit. The graph may show a vertical behavior suggesting a zero value or may need algebraic verification.

Analyze Indeterminate Form

Substitute \( x = 1 \) directly into the function. The expression becomes \( \frac{1-1}{1-\cos(\pi \times 1)} = \frac{0}{0} \), indicating an indeterminate form, which allows the application of L'Hôpital's Rule.

Apply L'Hôpital's Rule

With an indeterminate form \( \frac{0}{0} \), differentiate the numerator \( f(x) = x-1 \) to receive \( f'(x) = 1 \), and differentiate the denominator \( g(x) = 1 - \cos(\pi x) \) using the chain rule to receive \( g'(x) = \pi \sin(\pi x) \).

Evaluate the New Limit Expression

According to L'Hôpital's Rule, the limit becomes \( \lim_{x \rightarrow 1} \frac{f'(x)}{g'(x)} = \lim_{x \rightarrow 1} \frac{1}{\pi \sin(\pi x)} \). Substitute \( x = 1 \) into this expression to get \( \frac{1}{\pi \sin(\pi \times 1)} = \frac{1}{\pi \times 0} \).

Re-evaluate Using a Trigonometric Identity

Recognize that \( \sin(\pi x) \) can be approximated around \( x = 1 \). Utilize the limit property \( \lim_{z \rightarrow 0} \frac{\sin(z)}{z} = 1 \), rewriting the expression as \( \frac{1}{\pi (1-1)}\). This indicates continuous indeterminacy, revisit with trigonometric approximations or graph verification.

Conclude the Solution

Review graph and algebraic interpretation: Using trigonometric approximations, solving confirms the function has discontinuous abnormalities near \( x=1 \). Continuous zero behavior signifies the limit remains 0.

Key concepts

Indeterminate Forms

In calculus, an indeterminate form occurs when substituting a value into a function leads to an ambiguous situation, such as \( \frac{0}{0} \). This means you can't straightforwardly determine the limit of the function by direct substitution.
Recognizing this form allows us to apply specific mathematical techniques, such as L'Hôpital's Rule, which helps in resolving these forms by considering the rates of change (derivatives) of the numerator and the denominator.

• Zero over Zero (\( \frac{0}{0} \)): When both the numerator and the denominator approach zero, the form becomes undefined, which is the classic case for L'Hôpital's Rule.
• Infinity over Infinity (\( \frac{\infty}{\infty} \)): Another indeterminate form where both parts of the fraction approach infinity, indicating that L'Hôpital's Rule can be applied.

These forms require further analysis, often by using differentiation, to "break" the indeterminacy and find the limit more accurately.

Trigonometric Limits

Trigonometric limits involve understanding the behavior of functions that include trigonometric components like sine, cosine, or tangent as the variable approaches a particular value.
In our exercise, we encountered a trigonometric function in the denominator, \( 1 - \cos(\pi x) \). As \( x \) approaches 1, the behavior of this term becomes critical.
Trigonometric limits often leverage familiar limit properties:

• A fundamental property is: \( \lim_{z \rightarrow 0} \frac{\sin(z)}{z} = 1 \). This is incredibly useful when approximating limits involving trigonometric expressions.
• Other identities, such as \( \cos(x) = 1 - \frac{x^2}{2} + \cdots \) near small angles or specific values, can simplify the process.

Understanding these limits helps you resolve cases where trigonometric functions aren't directly substitutable, leading the way to simpler limit calculations.

Chain Rule Differentiation

The chain rule is essential for differentiating composite functions. It ensures you can find derivatives efficiently even when a function consists of other functions.
In the example exercise, the denominator \( g(x) = 1 - \cos(\pi x) \) required differentiation using the chain rule. This is due to its composition involving the cosine of another function, \( \pi x \).
Here's the basic idea of the chain rule step by step:

• Original Function: Identify the outside function and inside function. Here, the outside is \( -\cos(u) \) with \( u = \pi x \).
• Differentiation: Differentiate the outside function while keeping the inside unchanged, resulting in \( \sin(u) \).
• Multiply by the Derivative of the Inside: Multiply by the derivative of the inside function, which is \( \pi \) in this case since \( u = \pi x \).

This approach, when applied correctly, breaks down complex differentiation tasks into manageable steps, and it is fundamental in solving limits involving composite functions.