lim ( tan x) = ∞
x→π/2
True or false with full explanation
Taking into account that limx→π2sinx=sinπ2=1\lim\limits_{x\to\frac{\pi}{2}}\sin x=\sin\frac{\pi}{2}=1x→2πlimsinx=sin2π=1 and limx→π2cosx=cosπ2=0,\lim\limits_{x\to\frac{\pi}{2}}\cos x=\cos\frac{\pi}{2}=0,x→2πlimcosx=cos2π=0, we conclude that limx→π2tanx=limx→π2sinxcosx=∣10∣=∞.\lim\limits_{x\to\frac{\pi}{2}}\tan x=\lim\limits_{x\to\frac{\pi}{2}}\frac{\sin x}{\cos x}=|\frac{1}{0}|=\infty.x→2πlimtanx=x→2πlimcosxsinx=∣01∣=∞.
Answer: true
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