How do you find the limit of $\frac{2 x - 1}{| 2 x^{3} - x^{2} |}$ $(2x-1)/(abs(2x^3 - x^2))$ as x is approaching 0.5 from the negative side?

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How do you find the limit of $\frac{2 x - 1}{| 2 x^{3} - x^{2} |}$ $(2x-1)/(abs(2x^3 - x^2))$ as x is approaching 0.5 from the negative side?

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Answer 1 · 2015-03-06T03:31:00

Factor, then use definition and properties of the absolute value.

$\frac{2 x - 1}{| 2 x^{3} - x^{2} |} = \frac{2 x - 1}{| x^{2} | | 2 x - 1 |} = \frac{1}{x^{2}} \frac{2 x - 1}{| 2 x - 1 |}$ $(2x-1)/abs(2x^3-x^2)=(2x-1)/(abs(x^2)abs(2x-1))=1/x^2(2x-1)/abs(2x-1)$

As $x \to \frac{1}{2}$ $xrarr1/2$ (from either side), $\frac{1}{x^{2}} \to 4$ $1/x^2rarr4$

For $x < 0.5$ $x<0.5$ , we have $2 x < 1$ $2x<1$ and $2 x - 1 < 0$ $2x-1<0$ .

Hence, for $x < 0.5$ $x<0.5$ , we get

$| 2 x - 1 | = - (2 x - 1)$ $abs(2x-1)=-(2x-1)$ and

$\frac{2 x - 1}{| 2 x - 1 |} = \frac{2 x - 1}{- (2 x - 1)} = - 1$ $(2x-1)/abs(2x-1)=(2x-1)/-(2x-1)=-1$ .

Therefore, as $x \to \frac{1}{2}$ $xrarr1/2$ from the negative side, $\frac{2 x - 1}{| 2 x^{3} - x^{2} |} \to (4) (- 1) = - 4$ $(2x-1)/abs(2x^3-x^2)rarr(4)(-1)=-4$ .

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How do you find the limit of $\frac{2 x - 1}{| 2 x^{3} - x^{2} |}$ $(2x-1)/(abs(2x^3 - x^2))$ as x is approaching 0.5 from the negative side?

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