Proving Big oh of $(k^5 - k^3)$

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I am new to big oh notation and proofs and I can't really wrap my head around the proofs. I know how to prove that an expression is big Oh of a simple expression, lets say $k^2 \in O(k^5)$ but my brain just blanks out when I see negative numbers inside the brackets. Any hints or tips on how to prove the following:

$$165k^5 + k^2 \in O(k^5 - k^3)$$

Any help would be appreciated.

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Your best bet is to use the definition: $f(x) = O(g(x))$ if and only if there exist $M > 0$ and $x_0 > 0$ such that $|f(x)| \le Mg(x)$ for all $x \ge x_0$.

So, you need to find an $M$ and $x_0$ such that $|165x^5 + x^2| \le M(x^5 - x^3)$ whenever $x \ge x_0$. You can do this using a chain of inequalities. For example, when $x \ge 1$, we have $x^2 \le x^5$, so $165x^5+x^2 \le 165x^5 + x^5 = 166x^5$. This last value ($166x^5$) is simpler to work with; try to do the same thing with $x^5-x^3$.