Find the limit using       lim    sinx/x =1                                    x->0 lim 6x^2(cot3x)(csc2x) x->0Show step by step instructions

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to use the following trigonometric identities, such that:

`cot 3x = (cos 3x)/(sin 3x)`

`csc 2x = 1/(sin 2x)`

Hence, you need to evaluate the limit, such that:

`lim_(x->0) 6x^2*(cos 3x)/(sin 3x)*(1/(sin 2x))`

You need to split the limit, such that:

`lim_(x->0) 6cos 3x*lim_(x->0) x/(sin 3x)*lim_(x->0) x/(sin 2x)`

You need to create the remarcable limit `lim_(u(x)->0) (sin (u(x)))/(u(x)) = 1` such that:

`lim_(x->0) 6cos 3x*lim_(x->0) (3x)/(3sin 3x)*lim_(x->0) (2x)/(2sin 2x)`

Splitting the limit yields:

`lim_(x->0) 6*lim_(x->0) cos 3x*lim_(x->0) (1/3)*lim_(x->0) (3x)/(sin 3x)*lim_(x->0) (1/2)*lim_(x->0) (2x)/(sin 2x) = 6*(cos 0)*(1/3)*1*(1/2)*1`

`lim_(x->0) 6*lim_(x->0) cos 3x*lim_(x->0) (1/3)*lim_(x->0) (3x)/(sin 3x)*lim_(x->0) (1/2)*lim_(x->0) (2x)/(sin 2x)= 6*1*1/6*1*1`

Reducing duplicate factors, yields:

`lim_(x->0) 6*lim_(x->0) cos 3x*lim_(x->0) (1/3)*lim_(x->0) (3x)/(sin 3x)*lim_(x->0) (1/2)*lim_(x->0) (2x)/(sin 2x)= 1`

Hence, evaluating the given limit, using the identity `lim_(u(x)->0) (sin (u(x)))/(u(x)) = 1` , yields `lim_(x->0) 6x^2*cot 3x*csc 2x = 1.`

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