To approximate the function $\frac{\sin(x)}{x}$ using the Taylor series of $\sin(x)$ of degree 3 centered at $x = 0$, we can follow these steps:

Step 1: Write the Taylor series for $\sin(x)$ centered at $x = 0$

The Taylor series for $\sin(x)$ around $x = 0$ is given by:

$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$

If we only consider up to the third-degree term:

$\sin(x) \approx x - \frac{x^3}{6}$

Step 2: Divide by $x$ to approximate $\frac{\sin(x)}{x}$

Now, we approximate $\frac{\sin(x)}{x}$ by dividing the Taylor series of $\sin(x)$ by $x$:

$\frac{\sin(x)}{x} \approx \frac{x - \frac{x^3}{6}}{x}$

Simplifying this:

$\frac{\sin(x)}{x} \approx 1 - \frac{x^2}{6}$

Final Approximation

Thus, the approximation of $\frac{\sin(x)}{x}$ using the Taylor series of $\sin(x)$ of degree 3 centered at $x = 0$ is:

$\frac{\sin(x)}{x} \approx 1 - \frac{x^2}{6}$

This approximation is valid for values of $x$ close to 0.

Would you like any further details or have any questions?

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Here are 5 related questions you might find interesting:

1. How does the approximation change if we use a higher-degree Taylor series for $\sin(x)$?
2. What is the actual value of $\frac{\sin(x)}{x}$ at $x = 0.1$, and how does it compare to the approximation?
3. Can you derive a general formula for the Taylor series of $\frac{\sin(x)}{x}$ at $x = 0$?
4. What happens to the approximation as $x$ increases away from 0?
5. How can you use the remainder term to estimate the error in this approximation?

Tip: The approximation $\frac{\sin(x)}{x} \approx 1 - \frac{x^2}{6}$ is particularly useful for small values of $x$ in fields like physics and engineering, where such expressions often appear.