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Given that $$ lim_x to 0 dfracf(x)x $$ exists as a real number, I am trying to show that $lim_xto0f(x) = 0$. There is a similar question here:
Limit of f(x) knowing limit of f(x)/x.

But this question starts with the assumption of $$ lim_x to 0 dfracf(x)x = 0, $$ 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 $$ lim_x to 0 fracf(x)x = 0 $$ and then apply the product rule?

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
$$
lim_x to 0 f(x) = lim_x to 0 fracf(x)x times x = R times 0 = 0.
$$

We have that eventually

$$0le left|fracf(x)xright|le M$$

therefore

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

Let $lim_xto0dfracf(x)x=l$ then $bigg|dfracf(x)x-lbigg|leq M$ for some $Min mathbbR$. So $bigg|dfracf(x)xbigg|leq |l|+MRightarrow |f(x)|leq |x|(|l|+M) Rightarrow lim_xto 0 f(x)=0 $

Let $x_n rightarrow 0$.

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

$y_n rightarrow L.$

With

$f(x_n)=$

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

$lim_n rightarrow infty f(x_n)=$

$lim_n rightarrow infty((y_n)(x_n))=$

($lim_n rightarrow infty(y_n))(lim_n rightarrow infty(x_n))=$

$L cdot 0=0.$