Animated Solution for Mathematics - Limits, Continuity and Differentiability: If ${\lim }_{x\to 0}\frac{\cos (2x)+a\cos (4x)-b}{x^4}$ is finite, then $(a+b)$ is equal to :

When you see a problem like ${\lim }_{x\to 0}\frac{\cos (2x)+a\cos (4x)-b}{x^4}$, your first instinct might be to reach for L'Hopital's Rule. Stop. Take a breath. Differentiating a function four times is a siren song that leads to algebraic shipwreck.

Instead, we are going to use the Taylor series, the most powerful microscope in your mathematical toolkit. We know that the expansion for $\cos (\theta )$ is:

Because our denominator is $x^4$, we only care about the terms up to $x^4$. Anything higher will simply vanish into thin air as $x$ approaches zero.

Let us start by expanding $\cos (2x)$. By substituting $2x$ into our series, we get:

Now, let us tackle the second term, $a\cos (4x)$. Substituting $4x$ gives us:

Now, we bring them together into the numerator:

We group the terms by their powers of $x$:

Constant terms: $(1+a-b)$ $x^2$ terms: $-x^2(2+8a)$ * $x^4$ terms: $x^4(\frac{2}{3}+\frac{32a}{3})$

The full expression becomes:

If the constant term $(1+a-b)$ is not zero, the limit is $\infty$. Similarly, if the $x^2$ term is not zero, the limit is still $\infty$. For the limit to be finite, these coefficients must vanish.

We set the coefficients to zero:

From $2+8a=0$, we immediately find $a=-\frac{1}{4}$. Substituting this into $1+a-b=0$, we get $1-\frac{1}{4}-b=0$, which means $b=\frac{3}{4}$.

Finally, the question asks for $a+b$. Adding $-\frac{1}{4}$ and $\frac{3}{4}$ gives us:

The final answer is $\frac{1}{2}$. You have just mastered the art of Taylor series expansion. Keep this logic in your heart, and no limit will ever scare you again.