proving $(x^n)$ diverges for $x>1$ I got stocked to prove this fact $\lim_{n\to\infty} x^n=\infty$. I am assuming for a contradiction that $(x^n)$ convergence so for all $\epsilon>0$ there exists $N\ge 1$ such that $|x^n-l|\le\epsilon$. Then I am thinking if I choose $\epsilon=x-L$ but that would not work since we can not guarantee that $x<l$, So could anyone please help me with this I would appreciate that?

Assuming that $$ \lim_{n \to \infty} x^n = L $$ we must also have $$ xL = x \lim_{n \to \infty} x^n = \lim_{n \to \infty} x^{n+1} = L $$ so $xL = L$, so $x = 1$ or $L = 0$.

Now use our assumption that $x > 1$, so $L = 0$. But this is impossible, because $1 < x < x^2 < x^3 < \cdots$, implying that $|x^n - 0| > 1$ for all $n$.