Question

What is the value of the limit?

Answer 1

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

Explanation:

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