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I've tried many ways of solving this limit without using l'Hopital and I just can't figure it out. I know the answer is $3/2 \sin (2a).$

$$\lim_{x\,\to\,0} \frac{\sin(a+x)\sin(a+2x) - \sin^2(a)} x$$

Thank you!

mzp
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Andreina Ayala
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3 Answers3

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\begin{align*} &\dfrac{\sin(a+x)\sin(a+2x)-\sin^{2}a}{x}\\ &=2\cdot\dfrac{\sin(a+x)[\sin(a+2x)-\sin a]}{2x}+\dfrac{(\sin a)[\sin(a+x)-\sin a]}{x}\\ &\rightarrow 2\sin(a+0)(\sin x)'\bigg|_{x=a}+(\sin a)(\sin x)'\bigg|_{x=a}\\ &=2(\sin a)(\cos a)+(\sin a)(\cos a). \end{align*}

The true without L'Hopital: \begin{align*} &\dfrac{\sin(a+x)\sin(a+2x)-\sin^{2}a}{x}\\ &=\dfrac{(\sin a\cos x+\cos a\sin x)(\sin a\cos 2x+\cos a\sin 2x)-\sin^{2}a}{x}\\ &=\dfrac{\sin^{2}a\cos x\cos 2x+\cos^{2}a\sin x\sin 2x+\sin a\cos a(\sin 2x\cos x+\cos x\sin 2x)-\sin^{2}a}{x}\\ &=\dfrac{\sin^{2}a(\cos x\cos 2x-1)}{x}+\cos^{2}a\cdot\dfrac{\sin x}{x}\cdot\sin 2x+\dfrac{\sin 2a}{2}\cdot 3\cdot\dfrac{\sin 3x}{3x}, \end{align*} where \begin{align*} \dfrac{\cos x\cos 2x-1}{x}&=\dfrac{\cos x-1}{x}-2\cdot\dfrac{\sin x}{x}\cdot\cos x\\ &=\dfrac{-2\sin^{2}(x/2)}{x}-2\cdot\dfrac{\sin x}{x}\cdot\cos x\\ &\rightarrow 0 \end{align*}

user284331
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  • Literally, this is no L'Hopital. – user284331 Apr 06 '18 at 00:13
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    Actually, in a sense it still is, because you still need to show that the derivative of $\sin x$ is $\cos x$. – Martin Argerami Apr 06 '18 at 00:14
  • Ah... okay, if we are not allowed to use the derivative, then this answer does not address OP issue. – user284331 Apr 06 '18 at 00:16
  • Don't take it as criticism. These questions are never pure, and depend heavily on how the trigonometric functions are defined, and what properties are assumed. For instance, you could define sine and cosine using power series, and then this would be actually "no L'Hôpital"; but that's not the usual approach in the courses where these questions are asked (and requires knowledge about power series and differential equations). – Martin Argerami Apr 06 '18 at 00:18
  • To add to my comment, here is a way to construct sine and cosine from scratch. – Martin Argerami Apr 06 '18 at 00:24
  • That is superb. – user284331 Apr 06 '18 at 00:26
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Perhaps not the most elegant, but:

$\lim_\limits{x\to0} \frac{\sin(a+x)\sin(a+2x) - \sin^2(a)} x\\ \lim_\limits{x\to 0} \frac{\frac 12 (-\cos(2a + 3x) + cos x) - \sin^2(a)} x\\ \lim_\limits{x\to 0} \frac{\frac 12 (-\cos(2a)\cos 3x + \sin(2a)\sin (3x) + \cos x) - \sin^2(a)} x\\ \lim_\limits{x\to 0} \frac{\frac 12 (-\cos(2a)\cos 3x + \sin(2a)\sin (3x) + \cos x) - \frac 12 (1-\cos 2a)} x\\ \lim_\limits{x\to 0} \frac{-\cos(2a)\cos 3x + \sin(2a)\sin (3x) + \cos x - 1 +\cos 2a} {2x}\\ \lim_\limits{x\to 0} \frac{\cos(2a)(1-\cos 3x)}{2x} - \frac {1-\cos x}{2x} + \frac { \sin(2a)\sin (3x)}{2x}$

The first term evaluates to $0,$ the second term evaluates to $0,$ the third term evaluates to $\frac 32 \sin 2a$

Martin Argerami
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Doug M
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0

You could also have used Talor series around $x=0$ and even got more than the limit itself $$\sin(a+k x)=\sin (a)+k \cos (a)\,x-\frac{1}{2} k^2 \sin (a)\,x^2+O\left(x^3\right)$$ making $$\sin(a+ x)\sin(a+2 x)=\sin ^2(a)+3 \sin (a) \cos (a)\,x+ \left(2 \cos ^2(a)-\frac{5 }{2}\sin ^2(a)\right)\,x^2+O\left(x^3\right)$$ $$\frac{\sin(a+x)\sin(a+2x) - \sin^2(a)} x=3 \sin (a) \cos (a)\,+ \left(2 \cos ^2(a)-\frac{5 }{2}\sin ^2(a)\right)\,x+O\left(x^2\right)$$ which shows the limit and how it is approached.

If you simplify, you should notice that the end result is just $$\frac 32 \sin(2a)+\frac{1}{4} (9 \cos (2 a)-1)\,x+O\left(x^2\right)$$

Claude Leibovici
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