Evaluating a limit for a odd rooted radical

This is not a question which is much of an issue with the limit per se. Algebrically...I am kinda flat at this juncture...

$\frac{2+x}{\sqrt[5]{x^{5} -9}}$

My questions:-

1. Highest power of x, that I should take inside the radical? I believe it's 5 given the root of the radical is 5.

2. Like Square Root must be >=0 (as we don't consider complexies), should all the odd rooted radicals also be >=0 means $\sqrt[5]{x^{5} -9} >= 0$

3. Assuming the x is approaching a large negative value... (infinity is boring, but it could be the case...) How can I evaluate the limits of this radical as x approaches the said negativity? When calculated, I get positive 1 as the answer. Because power is odd, thus both numerator and denominator's sign get cancelled.

$\frac{\frac{2}{-x}+\frac{x}{-x}}{-(\sqrt[5]{\frac{x^{5}}{x^{5}} -\frac{9}{x^{5}}})}$

Given, $-(\frac{1}{x})^{5} = -(\frac{1}{x^{5}})$

• Now the biggest issue... What about if the function was like this? $\frac{2+x}{\sqrt[3]{x^{5} -9}}$
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(1) I'm not sure what you mean. (2) Odd roots have all real numbers as their domain. (3) As long as $x\neq\sqrt[5]9,$ the function is continuous, so just evaluate at the large negative value. As for the case when $x$ decreases without bound, note that for $x\neq\sqrt[5]9$ we have $$\frac{2+x}{\sqrt[5]{x^5-9}}=\cfrac{\frac2x+1}{\frac1x\sqrt[5]{x^5-9}}=\cfrac{\frac2x+1}{\sqrt[5]{1-\frac9{x^5}}},$$ so the function readily tends to $1$ as $x$ decreases (or increases) without bound.
As for your alternate function... (1) Still not sure what you mean. (2) Same answer as above. (3) For $x\neq\sqrt[5]9$ we have $$\frac{2+x}{\sqrt[3]{x^5-9}}=\cfrac{\frac2x+1}{\frac1x\sqrt[3]{x^5-9}}=\cfrac{\frac2x+1}{\sqrt[3]{x^2-\frac9{x^3}}},$$ so while the numerator on the far right tends to $1$ as $x$ decreases (or increases) without bound, the denominator increases without bound, and so the function tends to $0.$