The image shows a mathematical limit problem:

$\lim_{{x \to \frac{\pi}{4}}} \frac{\tan x - 1}{x - \frac{\pi}{4}}$

To evaluate this limit, we can use L'Hôpital's Rule since it results in an indeterminate form $\frac{0}{0}$.

Solution:

1. Differentiate the numerator and denominator separately:

   • Derivative of $\tan x - 1$ with respect to $x$ is $\sec^2 x$.
   • Derivative of $x - \frac{\pi}{4}$ with respect to $x$ is $1$.

2. Apply L'Hôpital's Rule: $\lim_{{x \to \frac{\pi}{4}}} \frac{\sec^2 x}{1} = \sec^2 \left( \frac{\pi}{4} \right)$

3. Evaluate $\sec^2 \left( \frac{\pi}{4} \right)$: $\sec \left( \frac{\pi}{4} \right) = \frac{1}{\cos \left( \frac{\pi}{4} \right)} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}$ $\sec^2 \left( \frac{\pi}{4} \right) = (\sqrt{2})^2 = 2$

Final Answer:

The limit is $2$. Therefore, the correct option is: $D. 2$

Would you like further explanation or have any questions about this?

Here are some additional questions for deeper understanding:

1. What is L'Hôpital's Rule and when can it be applied?
2. Why does the limit result in the indeterminate form $\frac{0}{0}$ in this problem?
3. How do you find the derivative of $\tan x$?
4. What are other ways to evaluate limits without using L'Hôpital's Rule?
5. How does the trigonometric identity $\sec^2 x = 1 + \tan^2 x$ relate to this problem?

Tip: Always check for indeterminate forms when evaluating limits to determine if applying L'Hôpital's Rule is appropriate.