If $$f(x)=\frac{\sin 3x+A\sin 2x+B\sin x}{x^5},$$ $x\neq 0$, is continuous at $x=0$, then find $A,B$ and $f(0)$. Do not use series expansion or L Hospital's rule.
As $f(x)$ is continuous at $x=0$,therefore its limit at $x=0$ should equal to its value.
Note that this question is to be solved without series expansion or L Hospital's rule,
I tried to find the limit $\lim_{x\to 0}\frac{\sin 3x+A\sin 2x+B\sin x}{x^5}$
$\lim_{x\to 0}\frac{\sin 3x+A\sin 2x+B\sin x}{x^5}=\lim_{x\to 0}\frac{3\sin x-4\sin^3x+2A\sin x\cos x+B\sin x}{x^5}=\lim_{x\to 0}\frac{3-4\sin^2x+2A\cos x+B}{x^4}\times\frac{\sin x}{x}$
$=\lim_{x\to 0}\frac{3-4\sin^2x+2A\cos x+B}{x^4}$
As the denominator is zero,so numerator has to be zero,in order the limit to be finite.
So, $3+2A+B=0. (1)$
I tried but I could not get the second equation between $A$ and $B$. I am stuck here. How do I continue?
Best Answer
Using trigonometric identities we have \begin{align} 3-4\sin^2 x+2A\cos x+B&=4(1-\sin^2 x)+2A\cos\left(2\cdot\frac{x}{2}\right)+B-1\\ &=4\cos^2\left(2\cdot\frac{x}{2}\right)+2A\cos\left(2\cdot\frac{x}{2}\right)+B-1\\ &=4\left(1-2\sin^2\frac{x}{2}\right)^2+2A\left(1-2\sin^2 \frac{x}{2}\right)+B-1\\ &=16\sin^4\frac{x}{2}-16\sin^2\frac{x}{2}-4A\sin^2\frac{x}{2}+2A+B+3\\ &=16\sin^4\frac{x}{2}-4(A+4)\sin^2\frac{x}{2}+2A+B+3\\ \end{align} In order to make the limit finite we must have $$A+4=0\quad\text{and}\quad 2A+B+3=0\qquad\iff\qquad \color{blue}{A=-4}\quad\text{and}\quad \color{blue}{B=5}$$ By taking those values we get \begin{align} \lim_{x\to 0}\frac{\sin 3x\color{blue}{-4}\sin 2x+\color{blue}{5}\sin x}{x^5}&=\left(\lim_{x\to 0}\frac{16\sin^4\frac{x}{2}}{x^4}\right)\left(\lim_{x\to 0}\frac{\sin x}{x}\right)\\ &=\left(\lim_{x\to 0}\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^4\left(1\right)\\ &=\color{blue}{1} \end{align} Since $f$ is continuous at $0$ it follows $f(0)=1$.