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?


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 $f(0)=1$.


Your first step is correct, you need $3+2A+B=0$ for the function to converge. Let's substitute $B=-3-2A$ into the original function. Them you have

$$ f(x)=\frac{\sin(3x)+A\sin(2x)+(-3-2A)\sin(x)}{x^5} $$

Using the angle sum formulas: $$ \sin(3x)=\sin(x)\cos(2x)+\cos(x)\sin(2x)=\sin(x)(\cos^2(x)-\sin^2(x))+2\sin(x)\cos(x)^2. $$ Now, using the standard trig identity, $\sin^2(x)=1-\cos^2(x)$. Therefore, $$ \sin(3x)=4\sin(x)\cos(x)^2-\sin(x) $$

Also using the angle sum formulas: $$ \sin(2x)=2\sin(x)\cos(x) $$

Substituting these into the original function, we have $$ f(x)=\frac{4\sin(x)\cos^2(x)-\sin(x)+2A\sin(x)\cos(x)-3\sin(x)-2A\sin(x)}{x^5}. $$ By combining the terms with $A$ and those without, we have $$ f(x)=\frac{4\sin(x)(\cos^2(x)-1)+2A\sin(x)(\cos(x)-1)}{x^5}=\frac{\sin(x)}{x}\cdot\frac{4(\cos^2(x)-1)+2A(\cos(x)-1)}{x^4} $$ Factoring a difference of squares, the numerator becomes $$ f(x)=\frac{\sin(x)}{x}\cdot\frac{4(\cos(x)-1)(\cos(x)+1)+2A(\cos(x)-1)}{x^4}$$ Factoring again, we have $$f(x)=\frac{\sin(x)}{x}\cdot\frac{\cos(x)-1}{x^2}\cdot\frac{4(\cos(x)+1)+2A}{x^2} $$

Observe that $$ \lim_{x\rightarrow 0}\frac{\cos(x)-1}{x^2}=\lim_{x\rightarrow 0}\frac{\cos(x)-1}{x^2}\cdot\frac{\cos(x)+1}{\cos(x)+1}=\lim_{x\rightarrow 0}\frac{\cos^2(x)-1}{x^2(\cos(x)+1)} $$ Then, using the standard trig identities, we get that this limit equals $$ \lim_{x\rightarrow 0}\frac{-\sin^2(x)}{x^2(\cos(x)+1)}=\lim_{x\rightarrow 0}\left(\frac{\sin(x)}{x}\right)^2\frac{-1}{\cos(x)+1}, $$ which we can see has a limit of $-\frac{1}{2}$.

Continuing from here, you also need the numerator to vanish for the limit to exist because the denominator vanishes. In other words, $\lim_{x\rightarrow 0}4(\cos(x)+1)+2A=8+2A$ must be $0$, so $A=-4$. After substitution, the entire formula becomes $$f(x)=\frac{\sin(x)}{x}\cdot\frac{\cos(x)-1}{x^2}\cdot\frac{4(\cos(x)-1)}{x^2} $$ Substituting the values that we've calculated the limit as $$ \lim_{x\rightarrow 0}f(x)=1\cdot\left(-\frac{1}{2}\right)\cdot\left(-\frac{4}{2}\right)=1. $$

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    $\begingroup$ You made a typographical error. After you write "factoring a difference of squares, the numerator becomes," you have a product of two quotients. The denominator of the second quotient should be $x^{4}$. $\endgroup$ – user74973 Dec 19 '15 at 15:21
  • $\begingroup$ @user74973 Thanks and corrected. $\endgroup$ – Michael Burr Dec 19 '15 at 16:14
  • $\begingroup$ I spied one other typographical error. In your expansion of sin(3x), you have "$\sin(x)^{2}$" instead of "$\sin^{2}x$." May I suggest that you use an algin environment in your equivalent expression for $f(x)$ after you write "by combining the terms with $A$ and those without, we have ..."? You "should" have the quotient $\sin{x}/x$ on the next line. (By the way, you may want to consider writing "$\sin{x}$" instead of "$\sin(x)$" and writing "$\cos{x}$" instead of "$\cos(x)$.") $\endgroup$ – user74973 Dec 19 '15 at 20:03
  • $\begingroup$ I should say that you have a nice argument. May I make a suggestion to make it more lucid? If you suppose the reader knows how to factor a difference of two squares and the value of the limit of $(1 - \cos{x})/x^{2}$ at 0, you can write consecutively seven equivalent expressions for $f(x)$ in a single align environment, and with the last expression, you can evaluate the limit of the function at 0. $\endgroup$ – user74973 Dec 19 '15 at 20:03
  • $\begingroup$ @user74973 Thanks for the suggestions. I'm on my phone right now, so I won't attempt the changes right away. $\endgroup$ – Michael Burr Dec 19 '15 at 20:24

\begin{align*} f(x) &= \frac{\sin(3x) + A\sin(2x) + B\sin{x}}{x^{5}} \\ &= \frac{(-4\sin^{3}x + 3\sin{x}) + 2A\sin{x}\cos{x} + B\sin{x}}{x^{5}} \\ &= \frac{\sin{x}}{x} \cdot \frac{-4\sin^{2}x + 3 + 2A\cos{x} + B}{x^{4}} \\ &= \frac{\sin{x}}{x} \cdot \frac{-4\sin^{2}x + 3 + 2A\cos{x} + B}{x^{4}} \\ &= \frac{\sin{x}}{x} \cdot \frac{4(1 - \sin^{2}x) - 1 + 2A\cos{x} + B}{x^{4}} \\ &= \frac{\sin{x}}{x} \cdot \frac{-4(1 - \cos^{2}x) + 3 + 2A\cos{x} + B}{x^{4}} . \end{align*} The given function is defined at $0$ if, and only if, $3 + 2A + B = 0$. So, $B = -3 - 2A$, and \begin{align*} f(x) &= \frac{\sin{x}}{x} \cdot \frac{-4(1 - \cos^{2}x) - 2A(1 - \cos{x})}{x^{4}} \\ &= \frac{\sin{x}}{x} \cdot \frac{1 - \cos{x}}{x^{2}} \cdot \frac{-4 - 2A - 4\cos{x}}{x^{2}} . \end{align*} Again, the given function is defined at $0$ if, and only if, $-4 - 2A -4 = 0$. So, $A = -4$, $B = 5$, and \begin{align*} f(x) &= \frac{\sin{x}}{x} \cdot \frac{1 - \cos{x}}{x^{2}} \cdot \frac{4(1 - \cos{x})}{x^{2}} \\ &= 4 \cdot \frac{\sin{x}}{x} \cdot \frac{1 - \cos{x}}{x^{2}} \cdot \frac{1 - \cos{x}}{x^{2}} . \end{align*} So, $\lim_{\scriptscriptstyle{x\to0}} f(x) = 1$.


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