$a=0$, $f(x)=x$, $g(x)=x^2$If $$ \lim\limits_{n \to \infty} \dfrac{f(n)}{g(n)} = \dfrac{a}{b}$$ with $a \ne b$, $b \ne 0$ then $$b f(n) \sim a g(n) $$ ?....
$\displaystyle \dfrac{b}{a} \lim_{n \to \infty} \dfrac{f(n)}{g(n)}= \dfrac{b}{a} \cdot \dfrac{a}{b}$If $$ \lim\limits_{n \to \infty} \dfrac{f(n)}{g(n)} = \dfrac{a}{b}$$ with $a \ne b$, $b \ne 0$ then $$b f(n) \sim a g(n) $$ ?....
Yes.. just in the asymptotic sense, the constants $a,b$ don't factor into the expression $f \sim g$ because they're negligible in the limit.$\displaystyle \dfrac{b}{a} \lim_{n \to \infty} \dfrac{f(n)}{g(n)}= \dfrac{b}{a} \cdot \dfrac{a}{b}$
$\displaystyle \lim_{n \to \infty} \dfrac{b \cdot f(n)}{a \cdot g(n)}= 1$
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