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**Mathstud28** You could do this proof by contradiction. Assume that there does not exist a $\displaystyle N$ such that $\displaystyle N\leqslant n\implies a_n>0$

By definition of $\displaystyle \lim_{n\to\infty}a_n=a$ we must have that for every $\displaystyle \varepsilon>0$ there exists a corresponding $\displaystyle N$ such that $\displaystyle N\leqslant n\implies|a_n-a|<\varepsilon$. So if we take $\displaystyle \varepsilon=\frac{a}{2}$ we must be able to find a $\displaystyle N$ such that $\displaystyle |a_n-a|<\frac{a}{2}$, but now by assumption $\displaystyle a\leqslant|a_n-a|$ which gives a contradiction.