When Can I Use That Rule?

Calculus Level 2

lim x 0 sin x 1 cos x = lim x 0 d d x ( sin x ) d d x ( 1 cos x ) ( 1 ) = lim x 0 cos x sin x ( 2 ) = lim x 0 d d x ( cos x ) d d x ( sin x ) ( 3 ) = lim x 0 sin x cos x ( 4 ) = sin 0 cos 0 = 0 ( 5 ) \begin{array} {l c l r } \displaystyle \lim_{x\to0} \dfrac{\sin x}{1-\cos x} &= &\displaystyle \lim_{x\to0} \dfrac{\frac d{dx}(\sin x)}{\frac d{dx} (1-\cos x)} & \quad (1) \\ \displaystyle &= &\displaystyle \lim_{x\to0} \dfrac{\cos x}{\sin x} & \quad (2) \\ \displaystyle &= &\displaystyle \lim_{x\to0} \dfrac{\frac d{dx}(\cos x)}{\frac d{dx}(\sin x)} & \quad (3) \\ \displaystyle &= &\displaystyle \lim_{x\to0} \dfrac{-\sin x}{\cos x} & \quad (4) \\ \displaystyle &= & -\dfrac{\sin0}{\cos 0} = 0 & \quad (5) \end{array}

The above is my attempt at proving that lim x 0 sin x 1 cos x = 0 \displaystyle \lim_{x\to0} \dfrac{\sin x}{1-\cos x} = 0 . In which step did I first commit a flaw in my logic?

Step 1 Step 2 Step 3 Step 4 Step 5

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1 solution

Hobart Pao
Aug 18, 2016

You're not allowed to use L'Hopital's when the numerator and denominator aren't both 0 0 or \infty . That occurs at Step 3 \boxed{\textbf{Step 3}}

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