Questions: lim as theta approaches 9 pi / 4 of (tan theta - 1) / (theta - 9 pi / 4) =

Solution Steps

To solve the limit problem, we can use L'Hôpital's Rule because the limit is in the indeterminate form \(\frac{0}{0}\). L'Hôpital's Rule states that if \(\lim_{x \to c} \frac{f(x)}{g(x)}\) is in the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then it can be evaluated as \(\lim_{x \to c} \frac{f'(x)}{g'(x)}\), provided the limit on the right-hand side exists.

1. Identify the functions \(f(\theta) = \tan \theta - 1\) and \(g(\theta) = \theta - \frac{9\pi}{4}\).
2. Compute the derivatives \(f'(\theta)\) and \(g'(\theta)\).
3. Apply L'Hôpital's Rule by taking the limit of the ratio of these derivatives as \(\theta\) approaches \(\frac{9\pi}{4}\).

We start with the limit expression: \[ \lim_{\theta \rightarrow \frac{9\pi}{4}} \frac{\tan \theta - 1}{\theta - \frac{9\pi}{4}}. \] Here, we define: \[ f(\theta) = \tan \theta - 1 \quad \text{and} \quad g(\theta) = \theta - \frac{9\pi}{4}. \]

Next, we compute the derivatives of \(f\) and \(g\): \[ f'(\theta) = \sec^2 \theta \quad \text{and} \quad g'(\theta) = 1. \] At \(\theta = \frac{9\pi}{4}\), we find: \[ f'(\frac{9\pi}{4}) = \sec^2\left(\frac{9\pi}{4}\right) = \tan^2\left(\frac{9\pi}{4}\right) + 1 = 2. \]

Since both \(f(\theta)\) and \(g(\theta)\) approach 0 as \(\theta\) approaches \(\frac{9\pi}{4}\), we can apply L'Hôpital's Rule: \[ \lim_{\theta \rightarrow \frac{9\pi}{4}} \frac{f'(\theta)}{g'(\theta)} = \lim_{\theta \rightarrow \frac{9\pi}{4}} \frac{2}{1} = 2. \]

Final Answer