Problem 17 - Calculus for Biology and Medicine

Question

Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow 0} \frac{3^{-x}-1}{2^{x}-1} $$

Step-by-Step Solution

Answer

The limit is \( -\frac{\ln(3)}{\ln(2)} \).

Step 1: Identify the Form of the Limit

First, substitute \( x = 0 \) into the function to determine its form. We calculate \( \frac{3^{-0} - 1}{2^{0} - 1} \), which results in \( \frac{0}{0} \). This is an indeterminate form, indicating that l'Hôpital's Rule may be applicable.

Step 2: Apply l'Hôpital's Rule

Since the limit results in an indeterminate form \( \frac{0}{0} \), apply l'Hôpital's Rule. Differentiate the numerator and the denominator separately. The derivative of the numerator \( 3^{-x} - 1 \) is \( -3^{-x} \ln(3) \), and the derivative of the denominator \( 2^{x} - 1 \) is \( 2^{x} \ln(2) \).

Step 3: Re-evaluate the Limit with Derivatives

Reconsider the limit using the derivatives calculated in Step 2. The new limit expression is \( \lim_{x \to 0} \frac{-3^{-x} \ln(3)}{2^{x} \ln(2)} \).

Step 4: Substitute x = 0 once more

Substitute \( x = 0 \) into the new limit expression. Calculate \( \frac{-3^{0} \ln(3)}{2^{0} \ln(2)} = \frac{-1 \cdot \ln(3)}{1 \cdot \ln(2)} \), resulting in \( \frac{-\ln(3)}{\ln(2)} \).

Step 5: Simplify the Result

Express the limit obtained in a clear format: The limit evaluates to \( -\frac{\ln(3)}{\ln(2)} \). This is the simplified outcome.

Key Concepts

Limits

Calculus is a fascinating branch of mathematics and at the heart of it lies the concept of limits. A limit essentially describes the value that a function approaches as the input approaches a certain point. In the exercise given, we are asked to find a limit as \( x \) approaches 0.

• The notation \( \lim_{x \to 0} \) means we are interested in seeing what happens to the function when \( x \) gets very close to 0.
• Limits are crucial for understanding concepts of continuity, derivatives, and integrals.
• They help in determining the behavior of functions at points where they aren't necessarily defined, and in understanding how functions behave at infinity.

Understanding limits is the stepping stone towards many other calculus concepts, as they allow us to quantify the idea of approximation, which is central to calculus.

l'Hôpital's Rule

l'Hôpital's Rule is an incredibly useful tool in calculus for evaluating limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule helps us simplify our calculation process to find the limit.First, let's identify when to actually use l'Hôpital's Rule:

• Check if the limit is resulting in an indeterminate form, like \( \frac{0}{0} \).
• If it is, differentiate the numerator and the denominator of the function separately.
• Then, evaluate the new limit formed by the derivatives.

In our problem, when we first plugged in \( x = 0 \), it yielded a \( \frac{0}{0} \) form, indicating the need for l'Hôpital's Rule.
After differentiating both the top and bottom expressions, we reevaluated the limit as \( \lim_{x \to 0} \frac{-3^{-x} \ln(3)}{2^{x} \ln(2)} \).
Finally, substituting back \( x = 0 \) results in \( -\frac{\ln(3)}{\ln(2)} \).
This simpler form shows the elegance and efficiency of l'Hôpital's Rule in solving tricky limit problems.

Indeterminate Forms

Indeterminate forms arise when calculating limits and the expression doesn't directly lend itself to a numerical value. Understanding these forms is essential to proper application of tools like l'Hôpital's Rule.There are various forms considered "indeterminate":

• \( \frac{0}{0} \) - Seen when both the numerator and denominator vanish as the variable approaches a certain value.
• \( \frac{\infty}{\infty} \) - Appears when both numerator and denominator grow infinitely.
• Other forms include \( 0^0 \), \( 1^\infty \), \( \infty - \infty \), etc.

Knowing whether an expression is indeterminate helps decide which calculus techniques to use.
In our exercise, discovering the \( \frac{0}{0} \) form signaled that straightforward substitution wasn't enough, thus guiding us to apply l'Hôpital's Rule.
Recognizing these forms is crucial for accurate problem-solving in calculus and enabling further exploration into the behavior of complex functions.