$$\lim_{k\to x}\frac{ \sin^2 k - \sin^2 x}{k - x}$$
I have tried to solve it over and over but couldnt.
I will be very happy if someone can show me the way to solve this .
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familair with limit definition of derivative ? – AgentS Sep 05 '14 at 15:11
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Thats basically it , the original question was to use that – user173959 Sep 05 '14 at 15:12
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I am stuck solving the limit tho , i need to solve it this way although i already know the answer will be 2sin x i need to understand the lim steps to solve this – user173959 Sep 05 '14 at 15:13
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Hint: use $$ \sin^2k-\sin^2x=(\sin k-\sin x)(\sin k+\sin x), \sin k-\sin x=2\sin\frac{k-x}{2}\cos\frac{k+x}{2}, \lim_{x\to 0}\frac{\sin x}{x}=1. $$
Added: Note \begin{eqnarray} \frac{\sin^2k-\sin^2x}{k-x}&=&\frac{(\sin k-\sin x)(\sin k+\sin x)}{k-x}\\ &=&\frac{2\sin\frac{k-x}{2}\cos\frac{k+x}{2}(\sin k+\sin x)}{k-x}\\ &=&\frac{2\sin\frac{k-x}{2}}{k-x}\cos\frac{k+x}{2}(\sin k+\sin x) \end{eqnarray} and now I think you can get the limit.
xpaul
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I dont understand can u elaborate more please ? With every step till we get 2 sin x? – user173959 Sep 05 '14 at 15:18
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You could also use Taylor series: since $\sin^2(y)=\frac{1-\cos(2y)}{2}$ and that the McLaurin series built at $ \cos(z)$ at $z=a$ is $$\cos(z)=\cos (a)-(y-a) \sin (a)-\frac{1}{2} (y-a)^2 \cos (a)+O\left((y-a)^3\right)$$ Now $$\lim_{k\to x}\frac{ \sin^2 (k) - \sin^2 (x)}{k - x}=\lim_{k\to x}\frac{ \cos^2( x) - \cos^2( k)}{k - x}=\lim_{k\to x}\frac{ \cos(2 x) - \cos(2 k)}{2(k - x)}$$ Using the development given above for $\cos(z)$ and replacing $z$ by $2k$ as well as $a$ by $2x$, you should arrive to $$\cos^2( x) - \cos^2( k)=2 (k-x) \sin (2 x)+2 (k-x)^2 \cos (2 x)+O\left((k-x)^3\right)$$
I am sure that you can take from here.
Claude Leibovici
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Using $\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B$:
$$\lim_{k\to x}\frac{\sin^2k-\sin^2x}{k-x}=\lim_{k\to x}\frac{\sin(k-x)}{k-x}\cdot\lim_{k\to x}\sin(k+x)$$
lab bhattacharjee
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