Saturday, November 25, 2017

Find the limits without L'Hôpital:limxto0fracxsinxxtanx=?

Find the limits without L'Hôpital's rule
limx0xsinxxtanx=?
My Try:
\lim_{ x \to 0 }\frac{\sin(\pi-x)-\sin x}{\tan(\pi+x)-\tan x}=?\\\lim_{ x \to 0 }\frac{2\sin(\frac{\pi}{2}-x)\cos(\frac{\pi}{2})}{\frac{\sin(\frac{π}{2}-x)}{\cos(\pi+x)\cos(x)}}=\lim_{ x \to 0 }\frac{(2\cos x)(-\cos x)(\cos(\frac{\pi}{2}))}{\cos x}=0
but:
\lim_{ x \to 0 }\frac{x-\sin x}{x-\tan x}=-1/2
Where is my mistake?

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