# Limit of $f(x)$ given that $f(x)/x$ is known

Given that $$\lim_{x \to 0} \dfrac{f(x)}{x}$$ exists as a real number, I am trying to show that $\lim_{x\to0}f(x) = 0$. There is a similar question here: Limit of f(x) knowing limit of f(x)/x.

But this question starts with the assumption of $$\lim_{x \to 0} \dfrac{f(x)}{x} = 0,$$ and all I am assuming is that the limit is some real number. So the product rule for limits doesn't really work here.

Or do I need to show that $$\lim_{x \to 0} \frac{f(x)}{x} = 0$$ and then apply the product rule?

• Try proving it by contradiction: what happens if $\lim_{x\to 0}f(x)\neq 0$? Commented Aug 23, 2018 at 9:00
• What prevents you from using product rule of limits? Perhaps you need to revisit the product rule in your text and then understand that it works fine here. Commented Aug 23, 2018 at 14:23
• @FurryFerretMan Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– user
Commented Sep 17, 2018 at 20:08

The product rule trick still works. If $\lim_{x \to 0} f(x)/x = R \in \mathbb R$, and obviously $\lim_{x \to 0} x = 0$, it follows that $$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{f(x)}{x} \times x = R \times 0 = 0.$$

• This is the most natural solution and I would not consider use of limit laws a trick rather it is the method. +1 Commented Aug 23, 2018 at 14:19

We have that eventually

$$0\le \left|\frac{f(x)}{x}\right|\le M$$

therefore

$$0\le \left|f(x)\right|\le M|x| \to 0$$

Let $\lim_{x\to0}\dfrac{f(x)}{x}=l$ then $\bigg|\dfrac{f(x)}{x}-l\bigg|\leq M$ for some $M\in \mathbb{R}$. So $\bigg|\dfrac{f(x)}{x}\bigg|\leq |l|+M\Rightarrow |f(x)|\leq |x|(|l|+M) \Rightarrow \lim_{x\to 0} f(x)=0$

Let $x_n \rightarrow 0$.

$y_n:= f(x_n)/x_n$, we have

$y_n \rightarrow L.$

With

$f(x_n)=$

$(f(x_n)/x_n)(x_n)=(y_n)(x_n)$.

$\lim_{n \rightarrow \infty }f(x_n)=$

$\lim_{n \rightarrow \infty}((y_n)(x_n))=$

($\lim_{n \rightarrow \infty}(y_n))(\lim_{n \rightarrow \infty}(x_n))=$

$L \cdot 0=0.$