If we have the following limit $\displaystyle\lim_{x \to -\infty} \frac{x}{\sqrt{x^2-x}}$ I understand that we have to divide the numerator by $x$ and the denominator by $-\sqrt{x^2}$ because x<0.
However, the book only mentions the case that we need to add a negative sign when we have to divide by $\sqrt{x^2}$. What about other cases? What is the general rule?
For example, if we need to divide (all when $x \to -\infty$):
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$\sqrt{x^4}$ Ex: $\displaystyle\lim_{x \to -\infty} \frac{x^2}{\sqrt{x^4-1}}$
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$\sqrt[3]{x^6}$ Ex: $\displaystyle\lim_{x \to -\infty} \frac{x^2}{\sqrt[3]{x^6-1}}$
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$\sqrt{x^6}$ Ex: $\displaystyle\lim_{x \to -\infty} \frac{x^3}{\sqrt{x^6-1}}$
Which of the above do we need to add a negative sign because of x<0?
Best Answer
Generally $$ \sqrt{x^{2n}}=|x^n|. $$ This is equal to $-(x^n)$ when $x<0$ and $n$ is odd.