If, x 0lim (2x)+a (4x)-bx^4is, finite,, then, (a+b), is, equal, to:

Question

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

Accepted Answer

To find the value of $a + b$ for which the following limit is finite:

$ \lim\limits_{x 	o 0} \frac{\cos(2x) + a \cos(4x) - b}{x^4} $

we start by expanding the cosine functions using their Taylor series:

$ \cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \ldots $

$ \cos(4x) = 1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} - \ldots $

Substitute these expansions into the limit expression:

$ \lim\limits_{x 	o 0} \frac{\left(1 - \frac{4x^2}{2!} + \frac{(2x)^4}{4!} - \ldots ight) + a\left(1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} - \ldotsight) - b}{x^4} $

To ensure that the limit is finite, the linear and quadratic terms must sum to zero. Let's equate the constant terms first:

For the limit to be finite:

$    1 + a - b = 0    $

Equating the coefficients of $x^2$:

$    -2 - 8a = 0    $

Solving the second equation, we have:

$ 8a = -2 \ a = \frac{-1}{4} $

Substitute $a$ back into the first equation to find $b$:

$ 1 + \frac{-1}{4} - b = 0 \ b = 1 - \frac{1}{4} \ b = \frac{3}{4} $

Thus, the sum of $a$ and $b$ is:

$ a + b = \frac{-1}{4} + \frac{3}{4} = \frac{1}{2} $

Therefore, the correct value of $a + b$ is $\frac{1}{2}$.

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