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How to prove that $n^{3.14}−2017n^{1.28}+1 \in ω(n^3)$ without using limit?

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    Hi Barbara, welcome to MSE. Could you tell us please what you have tried to prove this without using a limit? where did you get stuck? – 2017-01-18
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    @Daniel, Here is what I get stuck: $c*n^{0.57}+2017/n^{1.13} – 2017-01-18
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    @Barbara What is your definition for $\omega(\cdot)$? Normally you'd be looking for something like $n^{0.14} - 2017 n^{-1.72} + n^{-3} \ge c$. – 2017-01-18
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    @dxiv, here is what I did: $ cn^3 – 2017-01-18
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    What is $\omega (n^3)$? This is not standard notation. – 2017-01-18
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    @user254665 In the context, I assume that $\omega(\cdot)$ is the opposite of $o(\cdot)\,$, see [this](https://en.wikipedia.org/wiki/Big_O_notation#Family_of_Bachmann.E2.80.93Landau_notations) for example. The more standard notation would be $=\omega(n^3)$ rather than $\in \omega(n^3)$ though. – 2017-01-18

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Assuming the usual definition of $\omega(\cdot)\,$:

$n^{3.14}−2017n^{1.28}+1 = ω(n^3) \;\iff\; \forall c \gt 0 \; \exists N \gt 0 \;\big|\; n^{3.14}−2017n^{1.28}+1 \ge c\,n^3 \;\;\forall n \ge N$

The common, most direct way to prove that would be using limits, but since the OP specifically asked for an alternative proof without using limits, below is the outline of such a proof.

  • Divide by $n^3$ and write the condition as:

$$ n^{0.14}\left(1 − \frac{2017}{n^{1.72}}\right)+\frac{1}{n^3} \;\ge\; c $$

  • Find $N_1$ such that $1 − \frac{2017}{N_1^{1.72}} = \frac{1}{2}\,$, which is $N_1=\left(\frac{2017}{2}\right)^{1/1.72}\,$. By monotonicity of the LHS it follows that $1 − \frac{2017}{n^{1.72}} \ge \frac{1}{2}$ for $\forall n \ge N_1\;$ (1).   (Note that $\frac{1}{2}$ was an arbitrarily chosen constant, any other $\lambda \in (0,1)$ would have worked just as well for the rest of the argument.)

  • Find $N_2$ such that $N_2^{0.14} = 2c\,$, which is $N_2 = (2c)^{1/0.14}\,$. Again by monotonicity of the LHS it follows that $n^{0.14} \ge 2c\,$ for $\forall n \ge N_2\;$ (2).

  • Then for $\forall n \ge \max(N_1,N_2)\,$, by (1) and (2):

$$ n^{0.14}\left(1 − \frac{2017}{n^{1.72}}\right)+\frac{1}{n^3} \;\ge\; 2\,c \cdot\frac{1}{2} + \frac{1}{n^3} \;=\; c + \frac{1}{n^3} \;\gt\; c $$