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$$\lim_{x \to 0} \frac{\cos(x)-\log(1+x)-1+x}{\sin^2x} = ? $$

Please explain the steps also
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Answer is 0.

Apply L'Hospital's rule

$\lim_{x \to 0} \frac{\cos(x)-\log(1+x)-1+x}{\sin^2x} = \lim_{x \to 0} \frac{-\sin(x)-\frac{1}{1+x}+1}{2\sin x\cos x} $

$=\lim_{x \to 0} \frac{-1}{2\cos(x)}+\frac{x}{2\sin x\cos x}  = 0$
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Lim x tends to 0     (- sin x - 1/(1+x)  + 1) / 2 sinx cos x         ....0/0 case L'Hospital's rule applied

But it is still in 0/0 form .. So again differentiating

Lim x tends to 0     ( - cos x + 1/(1+x)^2 ) / 2 cos 2x           .. 2 sin x cos x = sin 2x

( Lim x tends to 0     - cos x  / 2 cos 2x  )  +    ( Lim x tends to 0     1/ 2 (1+x)^2 ) cos 2x )

Applying limits

= -1/2 + 1/2

= 0

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