We are tasked with solving the following limit:

$\lim_{x \to 0} \frac{\sin(2x) + x}{2x + 3\sin(4x)}$

Step-by-Step Solution:

1. Substitute $x = 0$ into the expression directly: $\frac{\sin(2(0)) + 0}{2(0) + 3\sin(4(0))} = \frac{0 + 0}{0 + 3(0)} = \frac{0}{0}$ Since this results in an indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule.

2. Differentiate the numerator and the denominator:

   • The numerator: $f(x) = \sin(2x) + x$ [ f'(x) = \frac{d}{dx}[\sin(2x)] + \frac{d}{dx}[x] = 2\cos(2x) + 1 ]
   • The denominator: $g(x) = 2x + 3\sin(4x)$ [ g'(x) = \frac{d}{dx}[2x] + \frac{d}{dx}[3\sin(4x)] = 2 + 12\cos(4x) ]

3. Evaluate the limit using the derivatives: $\lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{2\cos(2x) + 1}{2 + 12\cos(4x)}$ Substituting $x = 0$ into this expression: $\frac{2\cos(0) + 1}{2 + 12\cos(0)} = \frac{2(1) + 1}{2 + 12(1)} = \frac{3}{14}$

Thus, the value of the limit is:

$\boxed{\frac{3}{14}}$

Would you like further details on any step or have any other questions?

Relative questions:

1. How does L'Hopital's rule work for indeterminate forms like $\frac{0}{0}$?
2. What are some common indeterminate forms in calculus, and how are they resolved?
3. Can the Squeeze Theorem be applied to this problem instead of L'Hopital's Rule?
4. Why does direct substitution fail for some limits, leading to indeterminate forms?
5. How do trigonometric limits like $\lim_{x \to 0} \frac{\sin x}{x}$ play a role in solving limits?

Tip:

When using L'Hopital's Rule, always check if the limit is initially in an indeterminate form, as the rule applies only under those conditions.