https://file.aiquickdraw.com/mathbot/file/1725984655931urq3zew7.jpg

The expression in the image appears to be a limit problem involving trigonometric functions. It is written as:

\[
\lim_{{x \to \theta}} \frac{x \tan x - \theta \tan \theta}{x - \theta}
\]

This type of limit problem resembles a derivative definition using the difference quotient. To solve it, let's first recognize that the expression:

\[
\lim_{{x \to \theta}} \frac{f(x) - f(\theta)}{x - \theta}
\]

is the derivative of \( f(x) \) evaluated at \( x = \theta \). Therefore, we need to identify \( f(x) \). From the given expression, it seems that \( f(x) = x \tan x \).

### Step-by-step approach:

1. Recognize the expression as a derivative form:
   \[
   \lim_{{x \to \theta}} \frac{f(x) - f(\theta)}{x - \theta} = f'(\theta)
   \]
   Here, \( f(x) = x \tan x \).

2. Differentiate \( f(x) = x \tan x \) using the product rule:
   \[
   f'(x) = \frac{d}{dx} (x \tan x) = \tan x + x \sec^2 x
   \]

3. Substitute \( x = \theta \) into the derivative:
   \[
   f'(\theta) = \tan \theta + \theta \sec^2 \theta
   \]

Thus, the value of the limit is:

\[
\lim_{{x \to \theta}} \frac{x \tan x - \theta \tan \theta}{x - \theta} = \tan \theta + \theta \sec^2 \theta
\]

Let me know if you'd like any further details or clarification!

### Related Questions:
1. What is the product rule in differentiation, and how is it applied here?
2. How can we derive the derivative of trigonometric functions like \(\tan x\)?
3. Can this limit problem be solved using L'Hôpital's Rule? Why or why not?
4. What are the key differences between solving limits directly and using derivative definitions?
5. How do higher-order derivatives affect the evaluation of limits?

**Tip:** Recognizing the derivative form of a limit expression can often simplify problems involving complex trigonometric functions!

lim (x → θ) (x tan(x) - θ tan(θ)) / (x - θ)

Solve the limit lim (x → θ) (x tan(x) - θ tan(θ)) / (x - θ) using the concept of derivatives and trigonometric identities. This step-by-step solution covers limits and the product rule. Suitable for students in Grades 11-12.

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