# Calculate $\lim_{x \to 0} f(x)$

Let $$f:\mathbb{R}\to \mathbb{R}$$ be a function. Suppose $$\lim_{x \to 0} \frac{f(x)}{x} = 0$$. Calculate $$\lim_{x \to 0} f(x)$$.

According to the answer key, $$\lim_{x \to 0} f(x) = 0$$. I see $$f(x)=x^2$$ satisfies both limits, but is there a way "construct" an argument to prove this? I mean, how do I arrive at the answer?

• If $f(x) \not \to 0$ then $\frac {f(x)}x \to 0$ would be impossible. Commented Jul 7, 2020 at 23:47
• If $\frac{f(x)}{x}\to 0$, then for all $\varepsilon>0$, there exists a neighborhood of $x=0$ for which $$-\varepsilon x<f(x)<\varepsilon x$$ Commented Jul 8, 2020 at 3:03

$$\lim_{x \to 0} f(x)=\lim_{x \to 0} \frac {f(x)} x x =\lim_{x \to 0}\frac {f(x)} x \lim_{x \to 0}x=(0)(0)=0$$.
$$\lim_{x\to 0}\frac {f(x)}x = 0$$
For any $$\epsilon > 0$$ then there exists a $$\delta>0$$ so that if $$|x| < \delta$$ then $$|\frac{f(x)}x| < \epsilon$$. But if $$D = \min (\delta, 1)$$ then if $$|x| < D$$ then $$|f(x)|=|\frac {f(x)}1|< |\frac {f(x)}x| < \epsilon$$.
So $$\lim_{x\to 0}f(x) = 0$$.
Note: If $$\lim_{x\to 0} f(x) =k\ne 0$$ or if $$\lim_{x\to 0} f(x) = \pm \infty$$ then $$\lim_{x\to 0} \frac {f(x)}{x} = \frac{\lim_{x\to 0} f(x)}{\lim_{x\to 0} x}$$ is not defined and can not be $$0$$. But this allows the possibility the $$\lim f(x)$$ simply does not exist.