TITLE: Limit of f(x) knowing limit of f(x)/x
QUESTION [9 upvotes]: Given that:
$$\lim_{x\to0} \frac{f(x)}{x}=0$$
How can I prove that:
$$\lim_{x\to0} f(x)=0$$
?
Would the L'Hospital's Rule be applicable?

REPLY [2 votes]: This is not a fancy proof.  It is a grinding one from the epsilon-delta definition of limit suitable for an introductory calculus class.
The core of the idea is that if $f(x)$ is bounded away from $0$ as $x$ goes to zero, then $f(x)/x$ is also bounded away from zero, as for small $x$, dividing by $x$ makes things further from zero.
And things bounded away from zero don't have a limit of zero.

Suppose the limit as $x$ goes to zero of $f(x)$ is $a$, and $a > 0$.
Let $\epsilon = a/2$.  Then there is a $\delta$ such that for all $0 < x < \delta$, $a/2 < f(x) < 3a/3$.
Let $\delta_0$ be the minimum of $1/2$ and $\delta$.
(statement 1): Then for $0 < x < \delta_0$, $f(x)/x \geq f(x)*2 \geq 2a/2 = a$
If the limit of $f(x)/x$ as $x$ goes to zero is $0$, then let $\epsilon = a/2$.
(statement 2): Then there exists a $\delta_1 > 0$ such that for all $0 < x < \delta_1$, $-a/2 < f(x)/x < a/2$
Let $\delta_2$ be the least of $\delta_0$ and $\delta_1$.
For $x < \delta_2 <= \delta_1$, $f(x)/x < a/2$ by (2).
For $x < \delta_2 <= \delta_1$, $f(x)/x \geq a$ by (1).
Thus $a/2 > a$ or $1/2 > 1$, a contradiction.
A similar result holds of $a < 0$.  Spotting the places where I implicitly assumed $a$ was positive may be interesting.