We are tasked with evaluating:

$\lim_{x \to 0} \frac{\sin x}{x + \tan x}$

Step-by-Step Solution:

1. Substitute $x = 0$: Direct substitution leads to an indeterminate form:

   $\frac{\sin 0}{0 + \tan 0} = \frac{0}{0}$

   This is the indeterminate form $\frac{0}{0}$, so we can apply L'Hopital's Rule. L'Hopital's Rule states that if a limit is in the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, the limit can be rewritten as the limit of the derivatives of the numerator and the denominator.

2. Differentiate the numerator and denominator:

   • The derivative of $\sin x$ is $\cos x$.
   • The derivative of $x + \tan x$ is $1 + \sec^2 x$.

   So the limit becomes:

   $\lim_{x \to 0} \frac{\cos x}{1 + \sec^2 x}$

3. Substitute $x = 0$ into the new expression:

   • $\cos(0) = 1$
   • $\sec(0) = 1$, so $1 + \sec^2(0) = 1 + 1^2 = 2$

   Thus, the limit is:

   $\frac{1}{2}$

Final Answer:

$\lim_{x \to 0} \frac{\sin x}{x + \tan x} = \frac{1}{2}$

Would you like any further clarification or details on this solution?

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Here are 5 related questions for deeper understanding:

1. How do we determine when to apply L'Hopital's Rule?
2. What are other methods for solving indeterminate limits besides L'Hopital's Rule?
3. How do we differentiate more complex trigonometric functions?
4. What is the geometric interpretation of limits involving trigonometric functions?
5. How can we use series expansion to evaluate limits?

Tip: Always check if a limit results in an indeterminate form before applying L'Hopital's Rule.