Problem 11 - Thomas Calculus

Question

Use l'Hôpital's Rule to find the limits. \(\lim _{x \rightarrow \pi / 4} \frac{\sin x-\cos x}{x-\pi / 4}\)

Step-by-Step Solution

Answer

The limit is \(\sqrt{2}\).

Step 1: Verify Form is Indeterminate

First, substitute the limit point into the function. Substitute \(x = \frac{\pi}{4}\) into both the numerator \(\sin x - \cos x\) and the denominator \(x - \frac{\pi}{4}\). You will find that both result in \(0\). This confirms the form is \(\frac{0}{0}\), which means we can apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule

Since the limit presents an indeterminate form \(\frac{0}{0}\), differentiate the numerator and denominator separately with respect to \(x\). The derivative of the numerator, \(\sin x - \cos x\), is \(\cos x - (-\sin x) = \cos x + \sin x\). The derivative of the denominator \(x - \frac{\pi}{4}\) is \(1\). This gives a new limit to evaluate: \(\lim _{x \rightarrow \pi / 4} \frac{\cos x + \sin x}{1}\).

Step 3: Evaluate the New Limit

Now, evaluate the limit of the derivative: \(\lim _{x \rightarrow \pi / 4} (\cos x + \sin x)\). Substitute \(x = \frac{\pi}{4}\) into the expression. Since \(\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}\) and \(\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}\), it follows that \(\cos \frac{\pi}{4} + \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}\).

Key Concepts

Indeterminate Forms

When we discuss limits and calculus, an often-encountered situation is the indeterminate form. Indeterminate forms occur when directly substituting the limit point into the function causes ambiguity. This could lead us towards a result that isn't immediately clear, meaning their limits are undefined until further evaluated. Here are typical forms:

• \( \frac{0}{0} \)
• \( \frac{\infty}{\infty} \)
• \( 0 \cdot \infty \)
• \( \infty - \infty \)

For this particular exercise, when substituting \( x = \frac{\pi}{4} \) directly into the given function, the result is \( \frac{0}{0} \). That's why it's classified as an indeterminate form, allowing us to use tools like L'Hôpital's Rule to explore these limits further.
L'Hôpital's Rule becomes a valuable technique in simplifying these types of expressions, enabling us to further differentiate and evaluate them in search of clear solutions.

Differentiation

The process of differentiation is key in solving indeterminate forms using L'Hôpital's Rule. Differentiation in calculus refers to finding the derivative of a function, which describes how the function's outcome changes as its input changes. In practical terms, for any given function \( f(x) \), the derivative gives us the slope of the tangent to the function’s curve at any point.
In this exercise:

• The numerator \( \sin x - \cos x \) differentiates to \( \cos x + \sin x \), applying basic differentiation rules for trigonometric functions.
• The constant subtraction term in the denominator, \( x - \frac{\pi}{4} \), simplifies directly to 1.

Differentiating numerator and denominator gives us a new function where we can easily calculate the limit, avoiding the original undefined form. Differentiation here modifies the original problem into one more tractable, easing the path to finding the limit.

Limit Evaluation

Once we've appropriately differentiated the components of our fraction in the case of L'Hôpital's Rule, we proceed to evaluate the limit again. This process focuses on accurately assessing the behavior of the newly derived function as \( x \) approaches a specific value. First, examine the new limit function, \( \lim_{x \to \pi/4} \frac{\cos x + \sin x}{1} \). Here, the denominator no longer depends on \( x \), making the calculation straightforward as we only need to evaluate the numerator.

• Calculate \( \cos(\frac{\pi}{4}) \) which results in \( \frac{\sqrt{2}}{2} \).
• Similarly, \( \sin(\frac{\pi}{4}) \) also results in \( \frac{\sqrt{2}}{2} \).
• Adding these trigonometric values together, we find \( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \).

In essence, revised calculations using derivative functions make limit evaluation direct and clear. This process unveils the limit's logical consistency and correctness, transforming previous complexity into calculable answers.