To evaluate the limit

$\lim_{x \to 0} \frac{7x - \sin(x)}{x^2 + \sin(3x)},$

we start by analyzing the behavior of the numerator and the denominator as $x \to 0$.

Step 1: Evaluate the numerator and denominator at $x = 0$

• Numerator: $7x - \sin(x)$

  As $x \to 0$, both $7x$ and $\sin(x)$ approach 0. Thus, the numerator $7x - \sin(x) \to 0$.

• Denominator: $x^2 + \sin(3x)$

  As $x \to 0$, $x^2 \to 0$ and $\sin(3x) \to 0$. Therefore, the denominator $x^2 + \sin(3x) \to 0$.

Since both the numerator and the denominator approach 0, this is an indeterminate form of type $\frac{0}{0}$. We can apply L'Hôpital's rule to evaluate the limit.

Step 2: Apply L'Hôpital's Rule

L'Hôpital's rule states that if $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$, then:

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$

if this latter limit exists.

Compute the derivatives of the numerator and the denominator:

• Derivative of the numerator $f(x) = 7x - \sin(x)$:

$f'(x) = 7 - \cos(x)$

• Derivative of the denominator $g(x) = x^2 + \sin(3x)$:

$g'(x) = 2x + 3\cos(3x)$

Now apply L'Hôpital's rule:

$\lim_{x \to 0} \frac{7x - \sin(x)}{x^2 + \sin(3x)} = \lim_{x \to 0} \frac{7 - \cos(x)}{2x + 3\cos(3x)}$

Step 3: Evaluate the limit after applying L'Hôpital's rule

As $x \to 0$:

• $\cos(x) \to 1$, so $7 - \cos(x) \to 7 - 1 = 6$
• $2x \to 0$ and $\cos(3x) \to 1$, so $2x + 3\cos(3x) \to 0 + 3 \times 1 = 3$

Substitute these into the limit:

$\lim_{x \to 0} \frac{7 - \cos(x)}{2x + 3\cos(3x)} = \frac{6}{3} = 2$

Final Answer

$\boxed{2}$