لدينا العلاقة :
$\displaystyle{\displaylines{\forall x \in \mathbb{R} \quad \cos(2x) = \cos^{2}(x) - \sin^{2}(x)}}$$\displaystyle{\displaylines{\begin{array}{rcl}\cos(x) & = & \cos^{2}\left(\dfrac{x}{2}\right) - \sin^{2}\left(\dfrac{x}{2}\right) \\ \\& = & 1-2 \sin^{2}\left(\frac{x}{2}\right) \end{array}}}$
نعوض في النهاية :
$\displaystyle{\displaylines{ \begin{array}{rcl}\displaystyle\lim_{x \rightarrow 0} \dfrac{1 - \cos(x)}{x^2} & = & \displaystyle\lim_{x \rightarrow 0} \dfrac{2 \sin^{2}\left(\frac{x}{2}\right)}{x^2} \\ \\& = & \displaystyle\lim_{x \rightarrow 0} 2 \left( \dfrac{\sin(\frac{x}{2})}{x} \right)^2\end{array}}}$بوضع
$\displaystyle{\displaylines{X = \frac{x}{2}}}$لدينا
$\displaystyle{\displaylines{\lim_{x \rightarrow 0} \dfrac{\sin(\frac{x}{2})}{x} = \lim_{X \rightarrow 0} \dfrac{\sin(X)}{2 X} = \frac{1}{2}}}$بما أن الدالة
$\displaystyle{\displaylines{x\rightarrow x^2}}$ متصلة في النقطة
$\displaystyle{\displaylines{x = \frac{1}{2}}}$, لدينا :
$\displaystyle{\displaylines{\lim_{x \rightarrow 0} \left( \dfrac{\sin(\frac{x}{2})}{x} \right)^2 = \left( \lim_{x \rightarrow 0} \dfrac{\sin(\frac{x}{2})}{x} \right)^2 = \left( \dfrac{1}{2} \right)^2= \frac{1}{4}}}$إذن
$\displaystyle{\displaylines{\lim_{x \rightarrow 0} \dfrac{1 - \cos(x)}{x^2} = 2 \frac{1}{4} = \frac{1}{2}}}$
لدينا :
$\displaystyle{\displaylines{ \begin{array}{rcl}\displaystyle\lim_{x \rightarrow 0} \dfrac{e^{x^{2}} \cos(x) - 1}{x^2} & = & \displaystyle\lim_{x \rightarrow 0} \dfrac{e^{x^{2}} \cos(x) + e^{x^{2}} - e^{x^{2}}- 1}{x^2} \\& = & \displaystyle\lim_{x \rightarrow 0} \left( \dfrac{e^{x^{2}} (\cos(x) - 1)}{x^2} + \dfrac{e^{x^{2}} - 1}{x^2} \right)\end{array}}}$بوضع
$\displaystyle{\displaylines{X = x^2}}$ :
لدينا
$\displaystyle{\displaylines{\lim_{x \rightarrow 0} \dfrac{e^{x^{2}} - 1}{x^2} = \lim_{X \rightarrow 0^+} \dfrac{e^{X} - 1}{X} = 1}}$ولدينا
$\displaystyle{\displaylines{\lim_{x \rightarrow 0} e^{x^2} = 1}}$ و
$\displaystyle{\displaylines{\lim_{x \rightarrow 0} \dfrac{\cos(x) - 1}{x^2} = - \frac{1}{2}}}$إذن
$\displaystyle{\displaylines{\lim_{x \rightarrow 0} \dfrac{e^{x^{2}} \cos(x) - 1}{x^2} = - \frac{1}{2} + 1 = \frac{1}{2}}}$
لدينا صيغة Euler:
$\displaystyle{\displaylines{e^{ix} = \cos(x) + i \sin(x)}}$وبالتالي ومن أجل
$\displaystyle{\displaylines{n}}$ عدد صحيح طبيعي
$\displaystyle{\displaylines{e^{inx} = (\cos(x) + i \sin(x))^n = \cos(nx)+i \sin(nx)}}$
إذن :
$\displaystyle{\displaylines{\cos(nx) = \Re((\cos(x) + i \sin(x))^n)}}$* من أجل
$\displaystyle{\displaylines{n=2}}$ لدينا :
$\displaystyle{\displaylines{\begin{array}{rcl}\cos(2x) & = & \Re((\cos(x) + i \sin(x))^2) \\ \\& = & \Re(\cos(x)^2 - \sin(x)^2 + 2i \cos(x) \sin(x)) \\ \\& = & \cos(x)^2 - \sin(x)^2 \end{array}}}$
* من أجل
$\displaystyle{\displaylines{n=3}}$ لدينا :
$\displaystyle{\displaylines{\begin{array}{rcl}\cos(3x) & = & \Re((\cos(x) + i \sin(x))^3) \\ \\ & = & \Re \big( \cos(x)^3 + 3 i \cos(x)^2 \sin(x) \\ \\ & ~ & - 3 \cos(x) \sin(x)^2 - i \sin(x)^3 \big) \\ \\ & = & \cos(x)^3 - 3 \cos(x) \sin(x)^2 \\ \\ & = & \cos(x) (\cos(x)^2 - 3 \sin(x)^2)\end{array}}}$
إذن :
$\displaystyle{\displaylines{\begin{array}{rcl}f(x) & = & \cos(x) \cos(2x) \cos(3x) \\& = & \cos(x)^2 \times (\cos(x)^2 - \sin(x)^2) \\& ~ & \times (\cos(x)^2 - 3 \sin(x)^2)\end{array}}}$
بما ان
$\displaystyle{\displaylines{\cos(x)^2 = 1 - \sin(x)^2}}$لدينا
$\displaystyle{\displaylines{\begin{array}{rcl}\cos(x) \cos(2x) \cos(3x) & = & (1-\sin(x)^2) \times (1-2\sin(x)^2) \\& ~ & \times (1-4 \sin(x)^2)\end{array}}}$بعد التبسيط :
$\displaystyle{\displaylines{\cos(x) \cos(2x) \cos(3x) = 1 - 7 \sin(x)^2 + 14 \sin(x)^4 - 8 \sin(x)^6}}$
$\displaystyle{\displaylines{\begin{array}{rcl}\dfrac{1 - \cos(x) \cos(2x) \cos(3x)}{x^2} & = & 7 \left(\dfrac{\sin(x)}{x}\right)^2 - 14 \dfrac{\sin(x)^4}{x^2} \\ \\& ~ & + 8 \dfrac{\sin(x)^6}{x^2}\end{array}}}$
لدينا
$\displaystyle{\displaylines{\lim_{x \rightarrow 0} \frac{\sin^{2}(x)}{x^2} = \left( \lim_{x \rightarrow 0} \frac{\sin(x)}{x} \right)^2 = 1}}$ (لأن الدالة
$\displaystyle{\displaylines{x \rightarrow x^2}}$ متصلة في
$\displaystyle{\displaylines{1}}$).
$\displaystyle{\displaylines{\lim_{x \rightarrow 0} \frac{\sin(x)^4}{x^2} = \lim_{x \rightarrow 0} \frac{\sin(x)^6}{x^2} = 0}}$وبالتالي :
$\displaystyle{\displaylines{\lim_{x \rightarrow 0} \frac{1 - \cos(x) \cos(2x) \cos(3x)}{x^2} = 7}}$
