If $\lim\limits_{x \to 0} \frac{\cos(2x) + a\cos(4x) - b}{x^4}$ is finite, then $(a + b)$ is equal to:

Step 1: Analyze the limit form

For the limit to be finite when the denominator approaches zero ($x^4 \to 0$), the numerator must also approach zero. This is a necessary condition for L'Hopital's Rule to be applicable multiple times.

Step 2: Apply condition for numerator to be zero

Substitute $x=0$ into the numerator and set it equal to zero. Since $\cos(0) = 1$, we get an equation relating $a$ and $b$. This gives us the first condition for the limit to be finite.

Step 3: Use Maclaurin series expansion

To evaluate the limit, we can use the Maclaurin series expansion for $\cos(kx)$. We need terms up to $x^4$ because the denominator is $x^4$. The general expansion for $\cos(u)$ is $1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \dots$.

Step 4: Substitute series into the expression

Substitute the Maclaurin series for $\cos(2x)$ and $\cos(4x)$ into the numerator. Also, substitute $b = 1+a$ from our first condition. This allows us to simplify the expression and cancel out constant terms.

Step 5: Simplify the expression

After substituting $b=1+a$, the constant terms $1+a-b$ cancel out. We then group terms by powers of $x$. For the limit to be finite, the coefficients of $x^2$ in the numerator must also be zero, otherwise, the limit would be infinite.

Step 6: Set coefficient of $x^2$ to zero

For the limit to be finite, the coefficient of $x^2$ in the numerator must be zero. If it were not zero, the expression would simplify to a term like $C/x^2$, which would tend to infinity as $x \to 0$. Setting the coefficient to zero allows us to solve for $a$.

Step 7: Calculate $b$ and $a+b$

Now that we have the value of $a$, we can find $b$ using the relation $b = 1+a$. Finally, we calculate the sum $a+b$.

More from this chapter

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