$f(s) \approx g(s)$ if $\lim_{s \to \infty}\dfrac{f(s)}{g(s)} \neq 0$. Let $a, k,b > 0$ and $f(s) = a - \dfrac{k}{b + is} $. Why $$f(s) \approx \dfrac{1}{s} \ \ \text{as} \ \ s \to \infty$$
Note that $$ f(s) = a - \dfrac{k}{b + is} = \dfrac{(ab - k) + ias }{b + is}, $$ then $$ \lim_{s \to \infty}\dfrac{f(s)}{1/s} = \lim_{s \to \infty}\dfrac{s[(ab - k) + ias ]}{b + is} = \lim_{s \to \infty}\dfrac{[(ab - k) + ias ]}{\frac{b}{s} + i} = \infty $$ I can't :(. And I know $\lim_{s \to \infty} f(s) = a$?