evaluate the following limit
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Lt ( Cos x -1 + x²/2 ) / 3x²
x--->0
Lt ( Sin x + x ) / 6x
x--->0
Lt 1/6( Sin x / x + 1 )
x--->0
= 1/6
By using L'Hospitals rule
Lt ( Cos x -1 + x²/2 ) / 3x²
x--->0
Lt ( Sin x + x ) / 6x
x--->0
Lt 1/6( Sin x / x + 1 )
x--->0
= 1/6
By using L'Hospitals rule
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see here is the answer
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