Any other method- part 2

In contrary to the name of the question, I will be proceeding with the problem based on L'Hopital's Rule

If $f(x)$ is continuous at $x=0$ , then

$\large \displaystyle \lim_{x \to 0} f(x) = f(0)$

To evaluate $\displaystyle \lim_{x \to 0}$ , we will use L'Hopital's Rule

$\Rightarrow \displaystyle \lim_{x \to 0} \frac{A \cos x + Bx \sin x - 5}{x^4} = f(0)$

To apply L'Hopital's Rule, the limit should exist in the indeterminate form of $\frac{0}{0}$ .

So firstly, on applying the limit to the L.H.S, we get $A = 5$

$\displaystyle \lim_{x \to 0} \frac{A \cos x + Bx \sin x - 5}{x^4} = \lim_{x \to 0} \frac{-A \sin x + B \sin x + Bx \cos x}{4x^3} . . . . . . . . . \frac{0}{0} \text{form}$

$\displaystyle \Rightarrow \lim_{x \to 0} \frac{-A \cos x + B \cos x + B \cos x - Bx \sin x}{12x^2} . . . . . . .$ not of $\frac{0}{0} form$

So by applying $x \to 0$ to the numerator we get, $-A + 2B = 0 \Rightarrow B = \frac{5}{2}$

Now,

$\displaystyle \Rightarrow \lim_{x \to 0} \frac{-\frac{5}{2} \sin x}{12x} = -\frac{5}{24} \approx -0.20833$

Hence $f(0) = \boxed{-0.20833}$

My solution is as mentioned by Akshat Sharma .

Let $g(x) = A\cos x + B \sin x - 5$ . For $f(x)$ to be continuous at $x = 0$ . $g(x)$ must be of the form $g(x) = x^4 h(x)$ , so that $f(x) = \dfrac{x^4h(x)}{x^4} = h(x)$ , which is continuous at $x=0$ .

$\begin{aligned} g(x) & = A\blue{\cos x} + B\blue{\sin x} - 5 \quad \quad \small \color{#3D99F6}{\text{By Maclaurin series}} \\ & = A \blue{\left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ...\right)} + Bx \blue{\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ...\right)} - 5 \\ & = \red{A\left(1 - \frac{x^2}{2!}\right) + Bx^2 - 5} + A x^4 \left(\frac{1}{4!} - \frac{x^2}{6!} + \frac{x^4}{8!} - ...\right) - B x^4 \left(\frac{1}{3!} - \frac{x^2}{5!} + \frac{x^4}{7!} - ...\right) \end{aligned}$

To make $g(x) = x^4h(x)$ , we have:

$\begin{aligned} \color{#D61F06} {A\left(1 - \frac{x^2}{2!}\right) + Bx^2 - 5} & = 0 \quad \quad \small \color{#3D99F6}{\text{Equating the coefficients}} \end{aligned}$

$\Rightarrow \begin{cases} A - 5 = 0 & A = 5 \\ \dfrac{A}{2} - B = 0 & B = \dfrac{5}{2} \end{cases}$

Therefore, $f(0) = h(0) = \dfrac{A}{4!} - \dfrac{B}{3!} = \dfrac{5}{24} - \dfrac{5}{12} \approx \boxed{-0.208}$

Just use the expansion of sinx and cosx. The terms having power of x less than 4 should be cancelled to zero as the terms if not cancelled then they will result in 1/x, and since x tends to 0, 1/x will tend to infinity, and therefore value of limit will be coefficient of x^4 in the numerator as terms having power of x more than 4 would simply be equal to zero, as by dividing by x^4 would result in a term having x or x to the power something which would be equal to 0 as x is equal to 0.