x → 0 lim 1 − cos x sin x = = = = = x → 0 lim d x d ( 1 − cos x ) d x d ( sin x ) x → 0 lim sin x cos x x → 0 lim d x d ( sin x ) d x d ( cos x ) x → 0 lim cos x − sin x − cos 0 sin 0 = 0 ( 1 ) ( 2 ) ( 3 ) ( 4 ) ( 5 )
The above is my attempt at proving that x → 0 lim 1 − cos x sin x = 0 . In which step did I first commit a flaw in my logic?
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You're not allowed to use L'Hopital's when the numerator and denominator aren't both 0 or ∞ . That occurs at Step 3