Question

If K is a constant, show that K(x-a) = o(x-a) [yes that is a "o" not a "0"] if and only if K=0.

Answer 1

Question 1. If $K$ is a constant, show that $K(x - a) = o(x - a)$ if and only if $K = 0$.

Solution. The sufficiency is obvious, so prove the necessity. Recall that $f(x) = o(g(x))$ as $x \to a$ iff for any $\varepsilon > 0$ there is $\delta > 0$, such that $|f(x)| \leq \varepsilon |g(x)|$ for all $x$ with $0 < |x - a| < \delta$. Use this definition in the case when $f(x) = K(x - a)$ and $g(x) = x - a$: for any $\varepsilon > 0$ there is $\delta > 0$ such that $|K(x - a)| \leq \varepsilon |x - a|$ for all $x$, such that $0 < |x - a| < \delta$. Since $|x - a| > 0$, we can divide both sides of the above inequality by $|x - a|$ and obtain $|K| < \varepsilon$. But this is true for any $\varepsilon > 0$. Therefore, $|K| = 0$, i.e. $K = 0$.