limx→01−cosxx=0 \lim \limits_{x \to 0} \dfrac{1 - \cos x}{x} = 0 x→0limx1−cosx=0
limx→01−cosxx=limx→01−cosxx1+cosx1+cosx=limx→01−cos2xx(1+cosx)=limx→0sin2xx(1+cosx)=limx→0sinxxsinx1+cosx=limx→0sinxx⋅limx→0sinx1+cosx=1⋅02=0 \begin{align*} \lim \limits_{x \to 0} \dfrac{1 - \cos x}{x} &= \lim \limits_{x \to 0} \dfrac{1 - \cos x}{x} \dfrac{1 + \cos x}{1 + \cos x} \\ &= \lim \limits_{x \to 0} \dfrac{1 - \cos^{2} x}{x(1+\cos x)} \\ &= \lim \limits_{x \to 0} \dfrac{\sin^{2}x}{x(1+\cos x)} \\ &= \lim \limits_{x \to 0} \dfrac{\sin x}{x} \dfrac{\sin x}{1+\cos x} \\ &= \lim \limits_{x \to 0} \dfrac{\sin x}{x} \cdot \lim \limits_{x \to 0} \dfrac{\sin x}{1+\cos x} \\ &= 1 \cdot \dfrac{0}{2} \\ &= 0 \\ \end{align*} x→0limx1−cosx=x→0limx1−cosx1+cosx1+cosx=x→0limx(1+cosx)1−cos2x=x→0limx(1+cosx)sin2x=x→0limxsinx1+cosxsinx=x→0limxsinx⋅x→0lim1+cosxsinx=1⋅20=0
사인함수의 극한limx→0sinxx=1 \lim \limits_{x \to 0} \dfrac{\sin x}{x} = 1 x→0limxsinx=1
사인함수의 극한
limx→0sinxx=1 \lim \limits_{x \to 0} \dfrac{\sin x}{x} = 1 x→0limxsinx=1
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