\displaystyle\lim_{x\to 0}\dfrac{7x-\sin(x)}{x^2+\sin(3x)}

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

\[
\boxed{2}
\]

Learn how to evaluate the limit lim_{x \to 0} (7x - sin(x)) / (x^2 + sin(3x)) using L'Hôpital's Rule. This problem explores the behavior of trigonometric functions and the application of calculus concepts. Suitable for advanced high school students and calculus learners.

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