91Ó°ÊÓ

When encountering the computation of limits, one often stumbles upon indeterminate forms. These are fractions where both the numerator and denominator evaluate to zero or infinity when plugging in the limit value.
Examples of such forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), and similar forms like \( 0 \cdot \infty \) or \( \infty - \infty \). These forms tell us that direct substitution isn't going to work for evaluating limits.
This is exactly why we consider techniques like L'Hospital's Rule for further evaluation.

In our given problem, substituting \( x = 1 \) into \( \frac{1-x+\ln x}{1+\cos \pi x} \) yields \( \frac{0}{0} \), which is the classic indeterminate form. Identifying this allows us to proceed with L'Hospital's Rule to find the limit.

Limit evaluation is the process of finding the value that a function approaches as the input approaches a particular value. But sometimes, direct substitution won't work, especially when encountering indeterminate forms.

In such cases, we employ calculus-based methods to evaluate limits accurately. One popular and powerful technique we utilize is L'Hospital's Rule.

• This rule states that if the limit of \( \frac{f(x)}{g(x)} \) produces \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then \[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \] provided the limit on the right exists.

For our exercise, after translating the original limit to an indeterminate form \( \frac{0}{0} \), we apply L'Hospital's Rule, taking derivatives of both the numerator and the denominator.
This transforms the original hard-to-solve limit problem into simpler derivative-based calculations, which eventually lead us to a specific limit value.

Derivatives play a crucial role in calculus, particularly in evaluating limits when using L'Hospital's Rule. The derivative describes how a function changes as its input changes, giving the slope of the function at any point.

Applying L'Hospital's Rule requires computing derivatives of both the numerator and the denominator separately. For our specific problem:

• The numerator \( 1-x+\ln x \) has a derivative of \( -1 + \frac{1}{x} \), derived by recognizing that the derivative of \( x \) is \( 1 \) and that of \( \ln x \) is \( \frac{1}{x} \).
• The denominator \( 1+\cos \pi x \) differentiates to \( -\pi \sin \pi x \). Here, the chain rule is essential, multiplying the sinusoidal derivative by \( \pi \).

When these derivatives still result in an indeterminate form, we continue deriving until a calculable form emerges, as in our problem where a second derivative application was needed.