Key Concepts and Formulas

• Taylor Series Expansion of Cosine: The Taylor series expansion of $\cos(x)$ around $x=0$ is given by: $\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots$ This is crucial for analyzing limits involving trigonometric functions as $x \to 0$.
• Condition for Finite Limit: For a limit of the form $\lim_{x \to 0} \frac{f(x)}{x^n}$ to be finite, the numerator $f(x)$ must have a Taylor series expansion that starts with a term of at least $x^n$. This means all terms of degree less than $n$ in the numerator's expansion must be zero.

Step-by-Step Solution

Step 1: Analyze the Limit Expression We are given the limit: $L = \lim\limits_{x \to 0} \frac{\cos (2x) + a\cos (4x) - b}{x^4}$ For this limit to be finite, the numerator must tend to zero as $x \to 0$. This is because the denominator $x^4$ tends to zero. If the numerator did not tend to zero, the limit would be infinite.

Step 2: Apply Taylor Series Expansion to the Numerator We will use the Taylor series expansion for $\cos(u)$ around $u=0$, which is $\cos(u) = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \ldots$. For $\cos(2x)$, substitute $u=2x$: $\cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \ldots = 1 - \frac{4x^2}{2} + \frac{16x^4}{24} - \ldots = 1 - 2x^2 + \frac{2}{3}x^4 - \ldots$ For $\cos(4x)$, substitute $u=4x$: $\cos(4x) = 1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} - \ldots = 1 - \frac{16x^2}{2} + \frac{256x^4}{24} - \ldots = 1 - 8x^2 + \frac{32}{3}x^4 - \ldots$ Now, substitute these expansions into the numerator of the limit: $\text{Numerator} = \left(1 - 2x^2 + \frac{2}{3}x^4 - \ldots \right) + a\left(1 - 8x^2 + \frac{32}{3}x^4 - \ldots \right) - b$

Step 3: Equate Coefficients to Ensure Finite Limit For the limit $L$ to be finite as $x \to 0$ and the denominator is $x^4$, the terms in the numerator of degree less than 4 must cancel out.

• Constant Terms: Collect the constant terms in the numerator: $(1 + a - b)$ For the limit to be finite, this constant term must be zero: $1 + a - b = 0 \quad (*)$

• $x^2$ Terms: Collect the coefficients of $x^2$ in the numerator: $(-2 - 8a)x^2$ For the limit to be finite, this coefficient must also be zero: $-2 - 8a = 0$

Step 4: Solve for $a$ and $b$ From the equation for the $x^2$ terms: $-2 - 8a = 0$ $8a = -2$ $a = -\frac{2}{8} = -\frac{1}{4}$ Now, substitute the value of $a$ into equation $(*)$ to find $b$: $1 + a - b = 0$ $1 + \left(-\frac{1}{4}\right) - b = 0$ $1 - \frac{1}{4} - b = 0$ $\frac{3}{4} - b = 0$ $b = \frac{3}{4}$

Step 5: Calculate $a + b$ Now that we have found the values of $a$ and $b$, we can calculate their sum: $a + b = -\frac{1}{4} + \frac{3}{4}$ $a + b = \frac{-1 + 3}{4}$ $a + b = \frac{2}{4}$ $a + b = \frac{1}{2}$

Step 6: Verify the Finite Limit (Optional but Recommended) Let's check if the $x^4$ terms also allow for a finite limit. The coefficient of $x^4$ in the numerator is: $\frac{2}{3} + a\left(\frac{32}{3}\right)$ Substitute $a = -\frac{1}{4}$: $\frac{2}{3} + \left(-\frac{1}{4}\right)\left(\frac{32}{3}\right) = \frac{2}{3} - \frac{32}{12} = \frac{2}{3} - \frac{8}{3} = -\frac{6}{3} = -2$ So, the numerator up to the $x^4$ term is: $(1 + a - b) + (-2 - 8a)x^2 + \left(\frac{2}{3} + \frac{32a}{3}\right)x^4 + \ldots$ With $a = -1/4$ and $b = 3/4$, this becomes: $0 + 0x^2 + (-2)x^4 + \ldots$ The limit is then: $\lim\limits_{x \to 0} \frac{-2x^4 + \ldots}{x^4} = -2$ This confirms that the limit is indeed finite.

Common Mistakes & Tips

• Incorrect Taylor Series: Ensure you are using the correct Taylor series for $\cos(x)$ and substituting the arguments ($2x$, $4x$) accurately. Forgetting the factorial in the denominator or miscalculating powers is a common error.
• Missing Terms: When ensuring the limit is finite, systematically equate the coefficients of terms with powers less than the power in the denominator ($x^4$). Failing to equate the constant term or the $x^2$ term will lead to incorrect values of $a$ and $b$.
• Algebraic Errors: Be meticulous with your algebraic manipulations when solving for $a$ and $b$, and when calculating their sum.

Summary

To find the values of $a$ and $b$ that make the given limit finite, we used the Taylor series expansion of the cosine functions around $x=0$. For the limit $\lim_{x \to 0} \frac{f(x)}{x^4}$ to be finite, the numerator $f(x)$ must have its terms of degree less than 4 cancel out. By expanding $\cos(2x)$ and $\cos(4x)$ and equating the coefficients of the constant term and the $x^2$ term in the numerator to zero, we derived two linear equations in $a$ and $b$. Solving these equations yielded $a = -1/4$ and $b = 3/4$. The sum $a+b$ was then calculated to be $1/2$.

The final answer is $\boxed{\frac{1}{2}}$.