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To solve the limit:

\[
\lim_{x \to 0} \frac{\cos(x) - 1 + \frac{x^2}{2}}{14x^4}
\]

We are given a hint to use the Taylor series expansion for \(\cos(x)\). The Taylor series for \(\cos(x)\) about \(x = 0\) (Maclaurin series) is:

\[
\cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)
\]

### Step 1: Substitute the Taylor series for \(\cos(x)\)

Now, substituting this expansion for \(\cos(x)\) in the given limit expression:

\[
\lim_{x \to 0} \frac{\left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right) - 1 + \frac{x^2}{2}}{14x^4}
\]

### Step 2: Simplify the numerator

The terms \(1\) and \(-1\) cancel out, so the expression simplifies to:

\[
\lim_{x \to 0} \frac{- \frac{x^2}{2} + \frac{x^4}{24} + \frac{x^2}{2}}{14x^4}
\]

Next, observe that \(-\frac{x^2}{2} + \frac{x^2}{2} = 0\), leaving:

\[
\lim_{x \to 0} \frac{\frac{x^4}{24}}{14x^4}
\]

### Step 3: Simplify the fraction

This simplifies further:

\[
\frac{x^4}{24} \div 14x^4 = \frac{1}{24} \times \frac{1}{14} = \frac{1}{336}
\]

### Step 4: Final result

Since the limit is now a constant, the final result is:

\[
\boxed{\frac{1}{336}}
\]

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Would you like further clarification or more details on any part of this solution?

Here are 5 related questions you could explore:
1. What is the general form of the Taylor series for any trigonometric function?
2. How do you determine the error term in a Taylor series expansion?
3. Can the limit be solved using L'Hôpital's Rule? If so, how?
4. What other functions have similar Taylor expansions around zero?
5. How does the radius of convergence affect the use of the Taylor series?

**Tip**: Taylor series are extremely useful for approximating functions near a point, especially when solving limits or differential equations.

Evaluate lim (x->0) (cos(x) - 1 + x^2/2) / (14x^4). Hint: Use a Taylor series expansion for cos(x).

Solve the limit lim (x->0) (cos(x) - 1 + x^2/2) / (14x^4) using the Taylor series expansion for cos(x). This detailed solution covers the simplification process, substitution of the Taylor series, and the final result. Perfect for students in Grades 11-12 studying calculus.

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