Find a value of c that makes the following function continuous at xequals0. Explain why the value of c works.
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To make the function \( f(x) \) continuous at \( x = 0 \), we need to find a value of \( c \) such that the limit of \( f(x) \) as \( x \to 0 \) is equal to \( f(0) \), or:

\[
\lim_{{x \to 0}} f(x) = f(0) = c.
\]

For \( x \neq 0 \), the function is given by:

\[
f(x) = \frac{3x - \sin(3x)}{8x^3}.
\]

### Step 1: Compute the Limit as \( x \to 0 \)

To find \( \lim_{{x \to 0}} \frac{3x - \sin(3x)}{8x^3} \), we can use the Taylor expansion of \( \sin(3x) \) around \( x = 0 \):

\[
\sin(3x) \approx 3x - \frac{(3x)^3}{6} + O(x^5) = 3x - \frac{27x^3}{6} + O(x^5).
\]

Substituting this expansion into the expression for \( f(x) \):

\[
f(x) = \frac{3x - \left(3x - \frac{27x^3}{6} + O(x^5)\right)}{8x^3}.
\]

Simplify inside the numerator:

\[
f(x) = \frac{3x - 3x + \frac{27x^3}{6} - O(x^5)}{8x^3} = \frac{\frac{27x^3}{6} + O(x^5)}{8x^3}.
\]

Now, divide each term by \( 8x^3 \):

\[
f(x) = \frac{27}{48} + O(x^2) = \frac{9}{16} + O(x^2).
\]

As \( x \to 0 \), the higher-order terms \( O(x^2) \) go to zero, leaving us with:

\[
\lim_{{x \to 0}} f(x) = \frac{9}{16}.
\]

### Step 2: Set \( c = \frac{9}{16} \)

To make \( f(x) \) continuous at \( x = 0 \), we set \( f(0) = c \). Therefore:

\[
c = \frac{9}{16}.
\]

### Explanation

The value \( c = \frac{9}{16} \) works because it matches the limit of \( f(x) \) as \( x \to 0 \). By setting \( c = \frac{9}{16} \), we ensure that \( \lim_{{x \to 0}} f(x) = f(0) \), which makes \( f(x) \) continuous at \( x = 0 \).

Would you like a detailed breakdown of any step, or do you have further questions?

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Here are some related questions for further practice:

1. How do Taylor series help in finding limits of functions with indeterminate forms?
2. What is the importance of continuity at a point for piecewise functions?
3. Can you find a similar limit if \( \sin(3x) \) were replaced by \( \cos(3x) \)?
4. What would change if the power of \( x \) in the denominator was different, say \( x^2 \) instead of \( x^3 \)?
5. How can L'Hôpital's Rule be applied to this problem, if at all?

**Tip:** Always consider using Taylor expansions for trigonometric functions in limit problems involving small values of \( x \). This approach simplifies the expression and reveals key terms.

Find a value of c that makes the following function continuous at x = 0. Explain why the value of c works.

f(x) = {
  (3x - sin(3x)) / (8x^3), if x ≠ 0
  c, if x = 0
}

Explore how to find the value of c that ensures the continuity of a piecewise function at x = 0. This solution uses Taylor series expansion and limits, showing step-by-step how to set f(0) = c to match the limit of f(x) as x approaches zero. Perfect for undergraduate calculus students.

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