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The problem in the image is:

\[
\lim_{x \to 0} \frac{x - \sin(2x)}{2x + \tan(x)}
\]

To solve this limit as \(x \to 0\), we can use approximations of \(\sin(x)\) and \(\tan(x)\) for small values of \(x\), or apply L'Hopital's Rule if we obtain the indeterminate form \(\frac{0}{0}\).

### Step 1: Check if it's an indeterminate form
Substitute \(x = 0\) directly into the expression:

- \(\sin(0) = 0\), so \(x - \sin(2x) = 0 - \sin(0) = 0\).
- \(\tan(0) = 0\), so \(2x + \tan(x) = 0\).

Thus, we have \(\frac{0}{0}\), which is an indeterminate form, so we can apply **L'Hopital's Rule**.

### Step 2: Apply L'Hopital's Rule
L'Hopital's Rule states that for limits of the form \(\frac{0}{0}\), we can take the derivatives of the numerator and denominator:

\[
\lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)}
\]

#### Numerator:
\( f(x) = x - \sin(2x) \)

The derivative is:
\[
f'(x) = 1 - 2\cos(2x)
\]

#### Denominator:
\( g(x) = 2x + \tan(x) \)

The derivative is:
\[
g'(x) = 2 + \sec^2(x)
\]

### Step 3: Evaluate the limit
Now, we evaluate the limit of the new expression:

\[
\lim_{x \to 0} \frac{1 - 2\cos(2x)}{2 + \sec^2(x)}
\]

At \(x = 0\):
- \(\cos(0) = 1\), so \(1 - 2\cos(2(0)) = 1 - 2(1) = -1\).
- \(\sec^2(0) = 1\), so \(2 + \sec^2(0) = 2 + 1 = 3\).

Thus, the limit becomes:
\[
\frac{-1}{3}
\]

### Final Answer:
\[
\lim_{x \to 0} \frac{x - \sin(2x)}{2x + \tan(x)} = \frac{-1}{3}
\]

Would you like a detailed explanation or have any other questions? 

Here are 5 related questions for practice:
1. What is the limit of \(\frac{\sin(x)}{x}\) as \(x \to 0\)?
2. How do you apply L'Hopital's Rule for \(\frac{\sin(x)}{\tan(x)}\) as \(x \to 0\)?
3. What are the small angle approximations for \(\sin(x)\) and \(\tan(x)\)?
4. Can limits be computed without L'Hopital's Rule using series expansions?
5. How do you determine if a limit is an indeterminate form?

**Tip**: Always check if you are dealing with an indeterminate form before applying L'Hopital's Rule!

lim(x -> 0) [(x - sin(2x)) / (2x + tan(x))]

This limit problem involves the application of L'Hopital's Rule to solve an indeterminate form of type 0/0. Learn the step-by-step process, including using trigonometric small-angle approximations and differentiating the numerator and denominator.

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