Problem 6 - Calculus Single Variable

Question

Use l'Hôpital's Rule to find the limit, if it exists. \(\lim _{x \rightarrow 0} \frac{\cos (x)-1}{e^{x}-1}\)

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify Indeterminate Form

First, substitute \(x = 0\) into the given expression \(\frac{\cos(x) - 1}{e^x - 1}\) to see if we get an indeterminate form. We find \(\cos(0) - 1 = 1 - 1 = 0\) and \(e^0 - 1 = 1 - 1 = 0\), so our expression is \(\frac{0}{0}\), which is an indeterminate form. Thus, L'Hôpital's Rule can be applied.

Step 2: Differentiate the Numerator and Denominator

Differentiate the numerator and denominator separately with respect to \(x\). The derivative of the numerator \(\cos(x) - 1\) is \(-\sin(x)\). The derivative of the denominator \(e^x - 1\) is \(e^x\).

Step 3: Apply L'Hôpital's Rule

Now, apply L'Hôpital's Rule. The limit transforms to \(\lim_{x \to 0} \frac{-\sin(x)}{e^x}\).

Step 4: Evaluate the New Limit

Substitute \(x = 0\) into the new expression \(\frac{-\sin(x)}{e^x}\). This gives us \(\frac{-\sin(0)}{e^0} = \frac{0}{1} = 0\). Therefore, the limit is \(0\).

Key Concepts

Understanding Indeterminate Forms

The term "indeterminate forms" comes into play when evaluating limits in calculus. When you directly substitute a value into a function and encounter an expression like \( \frac{0}{0} \), you are dealing with an indeterminate form. This is because such an expression does not give a clear answer, as both the numerator and the denominator trend towards zero, causing ambiguity. Other common indeterminate forms include \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), and \( \infty - \infty \).
These expressions suggest that further manipulation is needed to properly evaluate the limit. In our exercise, substituting \( x = 0 \) into \( \frac{\cos(x) - 1}{e^x - 1} \) results in \( \frac{0}{0} \). Hence, we cannot simply determine the limit by substitution. Instead, we turn to tools like L'Hôpital's Rule to resolve it.

Introduction to Differentiation

Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. The derivative of a function at any point gives the slope of the tangent line to the function at that point. In simpler terms, it measures how the function's value changes as its input changes.
When applying L'Hôpital's Rule, differentiation is crucial. The rule states that if you have a limit resulting in an indeterminate form, you can differentiate the numerator and the denominator separately to find the limit of their quotient. In the exercise given, the derivative of \( \cos(x) - 1 \) is \( -\sin(x) \), and the derivative of \( e^x - 1 \) is \( e^x \). By differentiating both parts, we simplify the problem, making it possible to re-evaluate the limit.

Evaluating the Limit with L'Hôpital's Rule

Limit evaluation can sometimes be tricky, especially when dealing with indeterminate forms. L'Hôpital's Rule provides a technique to simplify and evaluate such limits by differentiating both the numerator and the denominator.
After differentiating \( \cos(x) - 1 \) and \( e^x - 1 \), we apply L'Hôpital's Rule to the limit \( \lim_{x \to 0} \frac{-\sin(x)}{e^x} \). By substituting \( x = 0 \), we get \( \frac{-\sin(0)}{e^0} \), resulting in \( \frac{0}{1} = 0 \).
This evaluation shows that the limit of the original indeterminate expression is \( 0 \). Hence, L'Hôpital's Rule is a powerful tool in calculus for resolving limits that initially appear undefined or complex.