calculus130
step by step using L'hopital rule:
limit of 1+cos(x)/x * sin(x) as x approaches π
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nubeer
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put pi in place of x. and see what you got.
nubeer
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hint
cos pi = -1
sin pi = 0
calculus130
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i get 0/0 if i plug pi with x
nubeer
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yes.. now apply the derivative in numerator and seperately in denominator.
calculus130
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i get -sinx/-xcosx
calculus130
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after plugging the ans is 0/-1
nubeer
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hmm please check the derivative in denominator again.. i think its wrong.. look you have to x in there so i believe you have to find derivative with product rule.
richyw
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to compute\(\frac{d}{dx}\left(x\sin x\right)\) you need to use the product rule.
richyw
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\[=x\cos x+\sin x\]
richyw
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\[=\lim_{x\rightarrow \pi}\frac{-\cos x}{\sin x +x\cos x}\]\[=\frac{0}{1+\pi\cdot 0}\]\[=0\]
calculus130
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thnx
nubeer
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i believe cos pi = -1
so how come got 0 in numerator.
richyw
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wow. big mistake haha
nubeer
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lol no worries.. :P happens sometimes with me too :P
richyw
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or wait, sorry I meant I had the wrong numerator the numberator should be sine
richyw
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\[\lim_{x\rightarrow \pi}\frac{-\sin x}{\sin x +x\cos x}\]Still works out to 0/1=0
richyw
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or actuall \(0/-\pi\)=0 haha.