Question

Find each limit. Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$ \lim _{x \rightarrow(1 / 2)^{-}} \frac{\ln (4-8 x)^{2}}{\tan \pi x} $$

Short answer

The limit is \(-\infty\).

Step-by-step solution

Check Indeterminate Form

First, let's check if the limit expression forms an indeterminate type when substituting \( x = \frac{1}{2}^- \). The numerator is \( \ln((4-8x)^2) \) and the denominator is \( \tan(\pi x) \). Substitute \( x = \frac{1}{2} \) into both the numerator and denominator:- Numerator: \( \ln((4-8 \times \frac{1}{2})^2) = \ln(0) \), which tends towards \(-\infty\).- Denominator: \( \tan(\pi \times \frac{1}{2}) = \tan(\frac{\pi}{2}) \), which tends towards \( \infty \).Therefore, the expression is an indeterminate form of \( \frac{-\infty}{\infty} \). We can apply l'Hôpital's Rule.

Differentiate Numerator and Denominator

Apply l'Hôpital's Rule, which states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} \) is an indeterminate form, then:\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]**Differentiate the Numerator:**\[ f(x) = \ln((4-8x)^2) \rightarrow f'(x) = \frac{1}{(4-8x)^2} \cdot 2(4-8x)(-8) \]After simplifying, \( f'(x) = \frac{-16}{4-8x} \).**Differentiate the Denominator:**\[ g(x) = \tan(\pi x) \rightarrow g'(x) = \sec^2(\pi x) \cdot \pi = \pi \sec^2(\pi x) \]

Apply l'Hôpital's Rule

Now substitute the derivatives back into the limit:\[ \lim_{x \to \frac{1}{2}^-} \frac{f'(x)}{g'(x)} = \lim_{x \to \frac{1}{2}^-} \frac{-16/(4-8x)}{\pi \sec^2(\pi x)} \]This simplifies to:\[ \lim_{x \to \frac{1}{2}^-} \frac{-16}{\pi \sec^2(\pi x) (4-8x)} \]

Evaluate the Limit

As \( x \to \frac{1}{2}^- \), the behavior of \( 4-8x \to 0^- \) ensures the numerator tends to \(-16/0^-\) which indicates the entire expression tends to a negative infinity (\(-\infty\)) overlap as denominator \(4-8x\) vanishes to zero with negative approach but limited by secant term maintaining bounded correction.

Key concepts

Indeterminate Forms

In calculus, an indeterminate form arises when you directly substitute a value into a limit, and the result appears undefined or ambiguous. Common examples include forms such as \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \((1^\infty)\), and more. These forms don't give us clear answers but indicate that deeper analysis is necessary.
In our problem, substituting \( x = \frac{1}{2}^{-} \) results in \(\frac{-\infty}{\infty}\).
This is an indeterminate form because we can't directly determine if the limit converges to a particular value. It's like having an answer that points in different directions. This is where l'Hôpital's Rule comes to the rescue, providing a method to evaluate such limits by differentiating the numerator and the denominator.

Differentiation

Differentiation is a fundamental concept in calculus that helps us understand how functions change. It's all about finding the derivative, which is the rate of change or slope of a function at a given point. With l'Hôpital's Rule, we use derivatives to simplify indeterminate forms.
For the given limit, we differentiate both the numerator and the denominator separately. The numerator is \( \ln((4-8x)^2) \). Through the chain rule, its derivative becomes \( \frac{-16}{4-8x} \). This tells us how the numerator behaves as \(x\) changes.
The denominator is \( \tan(\pi x) \), and its derivative is \( \pi \sec^2(\pi x) \). Differentiating it helps us see how the denominator changes. After differentiation, we can substitute these derivatives into the new limit expression.
This process ultimately lets us find the limit that was originally indeterminate.

Limits in Calculus

Limits are the foundation of calculus, allowing us to understand the behavior of functions as they approach particular points or values. They are crucial for defining both derivatives and integrals.
In this exercise, the limit \(\lim _{x \rightarrow(1 / 2)^{-}} \) signifies the behavior of the function as \(x\) approaches \(\frac{1}{2}\) from the left side. Since direct substitution led to an indeterminate form, using l'Hôpital's Rule became necessary.
After applying differentiation, we substitute back into the limit to finally evaluate it. As \( x \to \frac{1}{2}^- \), the unique behavior of both the numerator and denominator reveals that the limit tends towards \(-\infty\).
This showcases the power of combining indeterminate forms, differentiation, and limits to solve complex calculus problems efficiently.