Question
Asked Sep 22, 2019
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what is the limit of 

lim      xsinx / 2-2cosx
x--0

problems online show l'hopitals rule but i was wondering if there is another way to solve it without using that rule  

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Step 1

To determine the value of limit without using L hospital rule

xsin
limx0
2 2 cos.x
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xsin limx0 2 2 cos.x

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Step 2

Apply the limit o...

xsinx
limx0
2-2 cosx
rsinx(1cos.x
2-2 cos.x (1+cos x)
limx0
xsinx-(1cosx
limx0
2(1-cosx)(1+cosx)
limx0sinr-(1+cosx
2(1-cos'x)
limx0Sinr-(1+cosx)
2sin2 x
1-cos'xsin x
limr (1+ cos.x)
2sinx
x(1+cos.x)
limx0
sinx
2
limr +cos.r)
2
limx0
sinx
(1+cos0)
1+1
cos0-1}
2
2
(1+cos0)
= 1
2
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xsinx limx0 2-2 cosx rsinx(1cos.x 2-2 cos.x (1+cos x) limx0 xsinx-(1cosx limx0 2(1-cosx)(1+cosx) limx0sinr-(1+cosx 2(1-cos'x) limx0Sinr-(1+cosx) 2sin2 x 1-cos'xsin x limr (1+ cos.x) 2sinx x(1+cos.x) limx0 sinx 2 limr +cos.r) 2 limx0 sinx (1+cos0) 1+1 cos0-1} 2 2 (1+cos0) = 1 2

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Calculus

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