Can $a^2 = 2b^2$ have a solution where $a, b$ are in $\mathbb{Z}$ but not zero?
$\mathbb{Z}$ = positive and negative whole numbers
How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
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