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 $\lim_{x\to 0}f(x)\neq 0$? – Ender Wiggins Aug 23 '18 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. – Paramanand Singh Aug 23 '18 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 Sep 17 '18 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 – Paramanand Singh Aug 23 '18 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

$f(x_n)=$

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

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

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

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

$L \cdot 0=0.$