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

Please explain the steps also
in Calculus by Junior (985 points)
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+1
Apply L'Hospital's rule, it is easy after that.

3 Answers

+7 votes
Best answer
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} $

$=\lim_{x \to 0} \frac{-1}{2\cos(x)}+\frac{x}{2\sin x\cos x}  = 0$
by Boss (11.3k points)
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0

$\lim _{x\to \:0}\left(\frac{-1}{2\cos \left(x\right)}\right)=-\frac{1}{2}\\ \lim _{x \to 0}\frac{x}{2sinxcosx} = 0$

+1
$\lim_{x \to 0} \frac{x}{2\sin x\cos x} = \lim_{x \to 0} \frac{1}{2\frac{\sin x}{x}\cos x} = \frac{1}{2}$
0
Can you please explain the steps taken in simplification ?

From, $\lim_ {x \to 0}\frac{-\sin(x)-\frac{1}{1+x}-1+x }{2\sin(x)\cos(x)}$ to $\lim_ {x \to 0} \frac{-1}{2\cos(x)}+\frac{x}{2\sin(x)\cos(x)}$
+1
You have written the wrong expression, the correct one is (after applying L'Hospital rule) :

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

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

Yes there should be (1+x) in denominator as well, but that doesn't change the answer because that evaluates to 1 after applying limit.
0 votes

use : SIN^2 (X)=1/2(1-COS2X)

ANS: -2

by (83 points)
0 votes

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

by (21 points)
edited by

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