What is the value of the limit?

Let's multiply the top and bottom of the fraction by the conjugate #x+sqrtabsx# in order to get rid of that square root on the bottom.

#lim_(x->0)((x^2-2x)(x+sqrtabsx))/((x-sqrtabsx)(x+sqrtabsx))#

#= lim_(x->0)((x^2-2x)(x+sqrtabsx))/(x^2 - absx)#

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Now let's take the left and right limits. As #x# approaches #0# from the left (negative) side, #|x| = -x#. So we can say that:

#lim_(x->0^-)((x^2-2x)(x+sqrt(-x)))/(x^2 - (-x))#

#= lim_(x->0^-)((x^2-2x)(x+sqrt(-x)))/(x^2 + x)#

Now divide both sides by #x#. This puts the limit in a form where we can just plug in #0# and simplify.

#= lim_(x->0^-)((x-2)(x+sqrt(-x)))/(x+1)#

#= ((0-2)(0+sqrt0))/(0+1) = (-2(0))/1 = 0#

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As #x# approaches #0# from the right (positive) side, #|x| = x#. So we can say that:

#lim_(x->0^+)((x^2-2x)(x+sqrt(x)))/(x^2 - x)#

Now divide both sides by #x#. This puts the limit in a form where we can just plug in #0# and simplify.

#= lim_(x->0^+)((x-2)(x+sqrt(x)))/(x-1)#

#= ((0-2)(0+sqrt0))/(0-1) = (-2(0))/(-1) = 0#

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So #lim_(x->0^-)(x^2-2x)/(x-sqrtabsx) = 0# and #lim_(x->0^+)(x^2-2x)/(x-sqrtabsx) = 0#.

Since the left and right limit both equal zero, we can say that the limit itself is also equal to zero.

#lim_(x->0)(x^2-2x)/(x-sqrtabsx) = 0#

Final Answer