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Lim (x²-1) (2. Cosx)x_0 |
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Answer» L'Hopital's rules says that the lim x → a F ( x ) g ( x ) ⇒ f ' ( a ) g ' ( a ) Using this, we get lim x → 0 1 − cos x x 2 ⇒ − sin 0 2 ( 0 ) Yet as the DENOMINATOR is 0 , this is impossible. So we do a second LIMIT: lim ( x → 0 ) sin x 2 x ⇒ cos 0 2 = 1 2 = 0.5 So, in TOTAL lim x → 0 1 − cos x x 2 ⇒ lim x → 0 sin x 2 x ⇒ cos x 2 ⇒ cos 0 2 = 1 2 Step-by-step explanation: PLEASE MARK ME THE BRAINLEIST |
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