Problem 28 - Thomas Calculus in SI Units

Question

Use l'Hôpital's rule to find the limits. $$\lim _{\theta \rightarrow 0} \frac{(1 / 2)^{\theta}-1}{\theta}$$

Step-by-Step Solution

Answer

The limit is \(-\ln(2)\).

Step 1: Identify the Indeterminate Form

Recognize that the limit \( \lim_{\theta \to 0} \frac{(1/2)^\theta - 1}{\theta} \) is in the \( \frac{0}{0} \) indeterminate form because as \( \theta \to 0 \), both the numerator \((1/2)^\theta - 1\) and the denominator \(\theta\) approach 0.

Step 2: Apply l'Hôpital's Rule

Because the original limit forms an indeterminate type \( \frac{0}{0} \), we can apply l'Hôpital's Rule, which states that \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \), provided these derivatives exist.

Step 3: Differentiate the Numerator and Denominator

Differentiate the numerator \((1/2)^\theta - 1\) to get \( \frac{d}{d\theta}((1/2)^\theta) = (1/2)^\theta \ln(1/2) \). Differentiate the denominator \(\theta\) to get \(1\).

Step 4: Compute the New Limit

Substitute the derivatives into the limit: \( \lim_{\theta \to 0} \frac{(1/2)^\theta \ln(1/2)}{1} = \ln(1/2) \lim_{\theta \to 0} (1/2)^\theta \). As \( \theta \to 0 \), \((1/2)^\theta \to 1\). So, the expression simplifies to \( \ln(1/2) \cdot 1 = \ln(1/2) \).

Step 5: Simplify the Result

Recall that \( \ln(1/2) = -\ln(2) \). So, the limit \( \lim_{\theta \to 0} \frac{(1/2)^\theta - 1}{\theta} = -\ln(2) \).

Key Concepts

Limit Calculation

Calculating limits involves finding the value a function approaches as the input approaches a certain value. For example, with the function \( \lim_{\theta \to 0} \frac{(1/2)^\theta - 1}{\theta} \), we are interested in seeing what happens when \( \theta \) gets very close to 0. In situations like these, where direct substitution results in an undefined expression, we rely on clever methods to find the limit.One such method is l'Hôpital's Rule, which is particularly handy when limits contain indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). By differentiating the numerator and denominator separately, we transform the original limit into a simpler one that might be easier to evaluate. Once derivatives are computed, we substitute back into the limit and check the result. This process allows us to reveal the limiting behavior of the function. In our exercise, this technique neatly resolves the initial indeterminate form to an exact result.

Indeterminate Forms

Indeterminate forms occur when an algebraic expression evaluates to an ambiguous mathematical form. Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), and others. These forms don't provide clear answers, so we use additional tools to resolve them.In our example \( \lim_{\theta \to 0} \frac{(1/2)^\theta - 1}{\theta} \), both the numerator and the denominator approach 0, creating the \( \frac{0}{0} \) form. To resolve this, we use l'Hôpital's Rule which permits us to differentiate both the numerator and the denominator separately. The derivatives reveal the function's behavior as \( \theta \) approaches 0. This allows us to break down the indeterminate form and calculate the limit accurately.The ability to recognize and deal with indeterminate forms is crucial in calculus, especially when dealing with limits.

Exponential Functions

Exponential functions have the form \( f(x) = a^x \), where \( a \) is a constant and \( x \) is the exponent. These functions grow rapidly and are common in many areas of mathematics.In our exercise, we worked with \( (1/2)^\theta \), which is an example of an exponential function. As \( \theta \) approaches 0, \( (1/2)^\theta \) approaches 1.Exponential functions have unique properties, such as their rate of change being proportional to their current value. To differentiate an exponential function like \( f(\theta) = (1/2)^\theta \), we use the rule:\[ \frac{d}{d\theta}(a^\theta) = a^\theta \ln(a) \]With our base of \( 1/2 \), the derivative becomes \( (1/2)^\theta \ln(1/2) \). This knowledge allows us to apply l'Hôpital's Rule effectively in our exercise.Understanding exponential functions is vital as it enables us to calculate limits and work with various functions in calculus.