What is the value of the limit?

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Answer 1 · 2017-07-15T01:41:47

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

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