Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$\lim _{x \rightarrow 0}(\cos x-\sin x)^{1 / x}$$

Step-by-Step Solution

Answer

The limit is \(e\).

Step 1: Identify the Form of the Expression

Start by considering the expression \((\cos x - \sin x)^{1/x}\) as \(x\) approaches 0. Check if this creates an indeterminate form.First calculate \(\cos 0 - \sin 0\):\(\cos 0 = 1\) and \(\sin 0 = 0\), so \(\cos 0 - \sin 0 = 1\).This means the expression becomes \(1^{\infty}\) as \(x \to 0\), which is an indeterminate form.

Step 2: Apply Natural Logarithm

To tackle the \(1^{\infty}\) indeterminate form, consider taking the natural logarithm. Set \(y = (\cos x - \sin x)^{1/x}\). Then:\(\ln y = \frac{1}{x} \ln(\cos x - \sin x)\).

Step 3: Find the Limit of ln(y) using l'Hôpital's Rule

Now, evaluate \(\lim_{x \to 0} \frac{\ln(\cos x - \sin x)}{x}\). This gives us the form \(\frac{0}{0}\) as \(x \to 0\), so we can use l'Hôpital's Rule by differentiating the numerator and denominator separately.The derivative of the numerator \(\ln(\cos x - \sin x)\) is \(\frac{-\sin x - \cos x}{\cos x - \sin x}\).The derivative of the denominator \(x\) is 1.

Step 4: Continue Applying l'Hôpital's Rule

Thus, we apply l'Hôpital's Rule:\(\lim_{x \to 0} \frac{\ln(\cos x - \sin x)}{x} = \lim_{x \to 0} \frac{-\sin x - \cos x}{\cos x - \sin x}\).This still results in \(\frac{0}{0}\) as \(x \to 0\), so apply l'Hôpital's Rule again:

Step 5: Differentiate Again for l'Hôpital's Rule

Differentiate again with respect to \(x\).Numerator after differentiation: \(-\cos x + \sin x\).Denominator after differentiation: \(-\sin x - \cos x\.\)

Step 6: Evaluate the Limit

Now calculate \(\lim_{x \to 0} \frac{-\cos x + \sin x}{-\sin x - \cos x}\). Substitute \(x = 0\):This gives \(\frac{-1}{-1} = 1\).Thus, \(\lim_{x \to 0} \ln y = 1\).

Step 7: Solve for Final Limit

Recall \(y = e^{\ln y}\). Since \(\ln y = 1\), then \(y = e^1 = e\).Thus, the original limit \(\lim_{x \to 0}(\cos x - \sin x)^{1/x} = e\).

Key Concepts

Indeterminate Forms

When evaluating limits, you will often run into expressions that take on forms that are not straightforward to evaluate using standard limit rules. These are called indeterminate forms. Some common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 1^{\infty} \), \( \infty - \infty \), and \( 0 \times \infty \). In our example, \( \lim_{x \to 0} (\cos x - \sin x)^{1/x} \) presents the form \( 1^{\infty} \), which is indeed an indeterminate form. This indicates that direct substitution in the limit will not provide an answer, and we need to employ additional techniques to evaluate the limit effectively.

Limit Evaluation

Evaluating limits involving indeterminate forms often requires advanced techniques. One commonly used method is applying a logarithmic transformation. By rewriting the expression with natural logarithms, you can often reveal simpler forms that allow for classical limit evaluation techniques.
In the example at hand, dealing with the indeterminate form of \( 1^{\infty} \), we take the natural log of the expression \( y = (\cos x - \sin x)^{1/x} \) leading to \( \ln y = \frac{1}{x} \ln(\cos x - \sin x) \). This expression then allows us to focus on finding the limit of the transformed, simpler expression instead.

Calculus Concepts

Using calculus, particularly the technique known as L'Hôpital's Rule, can simplify the process of evaluating difficult limits stemming from indeterminate forms. L'Hôpital's Rule states that for limits resulting in the indeterminate forms \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), you can differentiate the numerator and the denominator separately to find the limit.
In this problem, once logs are applied, the expression becomes \( \lim_{x \to 0} \frac{\ln(\cos x - \sin x)}{x} \), which simplifies to \( \frac{0}{0} \), an appropriate situation for using L'Hôpital's Rule.
This involves differentiating both parts, leading to an expression whose limit can be evaluated more straightforwardly after just a couple of rounds of applying L'Hôpital's Rule. The final limit, found in terms of the logarithmic transformation \( \ln y \), then helps you determine the original limit in simple terms.