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Prove that $$ n^2 \cdot (1 + o(1)) = \Omega(n^2) $$

I'm having problems interpreting the LHS of this equation, since we are multiplying a function with little oh.

Bhishmaraj
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    Think of the $o(1)$ as a concrete function $f(n)$, which has the property that $\lim_{n\to\infty} f(n)/n=0$. Then prove that $C_1 n^2\le n^2(1+f(n))\le C_2 n^2$ for some constants $C_1,C_2$ (and large enough $n$). Hint: you can have $C_{1,2}=1\pm \epsilon$ for any $\epsilon>0$. – Mike Earnest Sep 20 '18 at 16:57
  • Thanks a lot ! Got it now . – Bhishmaraj Sep 20 '18 at 17:10

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