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Question
Using L'Hospital's rule, evaluate : `lim_(x->0) (x - sinx)/(x^2 sinx)`
Solution
`lim_(x->0) (x-sinx)/(x^2 sinx)`
`= lim_(x-> 0) (1-cos x)/(x^2 cos x + sin x. 2x)`
`= lim_(x->0) (2sin^2 x/2)/(x^2(cos x + (2sinx)/x)) = (2sin^2 x/2)/(4xxx^2/4(cos x + (2sinx)/x))`'
`= lim_(x- > ) 2/(4(cosx + 2))`
`= lim_(x -> 0) 1/(2(3)) = 1/6`
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