The image you uploaded contains two main tasks:

1. The point $P(x, y) = \left( \frac{1}{3}, \ln\left( \frac{1}{3} \right) \right)$ is to be considered, and you need to use linear approximation to estimate the value of the natural logarithm $\ln(x)$ at a point close to $\frac{1}{3}$, which is most likely $x = 0.4$.

2. Using the slope table: You are also asked to find the slope $m$ of the tangent lines to the graph of $f(x) = \ln(x)$ near $x = \frac{1}{3}$ and use that information for linear approximation.

To tackle these tasks:

Task 1: Use linear approximation formula.

The linear approximation (or tangent line approximation) formula for a function $f(x)$ at $x = a$ is:

$L(x) = f(a) + f'(a)(x - a)$

For $f(x) = \ln(x)$, we know that $f'(x) = \frac{1}{x}$. Therefore, you need to calculate $f\left( \frac{1}{3} \right) = \ln\left( \frac{1}{3} \right)$ and $f'\left( \frac{1}{3} \right) = \frac{1}{\frac{1}{3}} = 3$.

Then, using the linear approximation formula to estimate $\ln(0.4)$, we substitute into the equation.

Task 2: Slope table.

From the slope table in the image, you can use the slope values for $x$ close to $\frac{1}{3}$ and compare them to the theoretical slope of 3, which was calculated earlier.

Would you like me to go through the detailed calculation and provide an estimated value for $\ln(0.4)$ using this method?

Here are 5 related questions to deepen your understanding:

1. How is the derivative of $\ln(x)$ generally derived?
2. What is the geometric meaning of linear approximation?
3. Why is it important to choose a point $x = a$ close to the point you want to approximate?
4. Can you apply linear approximation to other functions, such as trigonometric ones?
5. How does the slope of a tangent line affect the accuracy of an approximation?

Tip: Linear approximations are particularly useful for estimating values of functions near points where you know the exact function and its derivative.