The graph of the function $g(x) = \frac{4x + x^3 \sin(2\pi x + \pi)}{6x}$ over the interval $[-2\pi, 2\pi]$ is shown above.

Step 2: Estimate the Limit from the Graph

From the graph, as $x$ approaches 0, the value of $g(x)$ appears to approach $\frac{2}{3}$.

Step 3: Justify the Limit Algebraically

To find the limit algebraically, we calculate:

$\lim_{x \to 0} g(x) = \lim_{x \to 0} \frac{4x + x^3 \sin(2\pi x + \pi)}{6x}$

We can separate the limit into two parts:

$\lim_{x \to 0} g(x) = \lim_{x \to 0} \left(\frac{4x}{6x} + \frac{x^3 \sin(2\pi x + \pi)}{6x}\right)$

Simplifying each term:

$\lim_{x \to 0} \frac{4x}{6x} = \frac{4}{6} = \frac{2}{3}$

For the second term:

$\lim_{x \to 0} \frac{x^3 \sin(2\pi x + \pi)}{6x} = \lim_{x \to 0} \frac{x^2 \sin(2\pi x + \pi)}{6}$

As $x \to 0$, $\sin(2\pi x + \pi)$ is bounded, so $x^2 \sin(2\pi x + \pi)$ approaches 0. Therefore:

$\lim_{x \to 0} \frac{x^2 \sin(2\pi x + \pi)}{6} = 0$

Thus, the overall limit is:

$\lim_{x \to 0} g(x) = \frac{2}{3} + 0 = \frac{2}{3}$

This confirms the estimate from the graph.

Would you like more details or have any questions?

Here are 5 related questions:

1. How do limits behave when dealing with trigonometric functions?
2. What is L'Hôpital's Rule, and when can it be used for limits?
3. How do you determine the continuity of a function at a point?
4. What role does the squeeze theorem play in evaluating limits?
5. How can you verify a limit algebraically when the function involves polynomials and trigonometric terms?

Tip: When estimating limits from a graph, always check for continuity and smooth behavior around the point of interest to ensure accuracy.