Q:

If $\lim_{x \rightarrow 0}\frac{f(x)-3e^{-x}}{x}=9$ what is $\lim_{x \rightarrow 0}f(x)$?

I'm trying to find $\lim_{x \rightarrow 0}f(x)$ given $\lim_{x \rightarrow 0}\frac{f(x)-3e^{-x}}{x}=9$. 
The way I did the question was to do 
\begin{align*}
& \frac{\lim_{x \rightarrow 0}f(x) - \lim_{x \rightarrow 0}3e^{-x}}{\lim_{x \rightarrow 0}x}=9 \\
&\rightarrow \frac{\lim_{x \rightarrow 0}f(x) - 3}{\lim_{x \rightarrow 0}x}=9 \\
&\rightarrow \lim_{x \rightarrow 0}f(x) - 3=9\lim_{x \rightarrow 0}x \\
&\rightarrow \lim_{x \rightarrow 0}f(x) - 3=9.0 \\
&\rightarrow \lim_{x \rightarrow 0}f(x) - 3=0 \\
&\rightarrow \lim_{x \rightarrow 0}f(x)=9
\end{align*}
which is the correct answer. However, I'm not sure if I'm allowed to multiply by $\lim_{x \rightarrow 0}x$ since it's equal to $0$.

A:

Note that
$$
\lim_{x\to0} (f(x)-3e^{-x})
=\lim_{x\to0}\left(\frac{f(x)-3e^{-x}}{x}\right)\times\lim_{x\to0}x
=9\times 0=0
$$
since both of  the limits in the product exist. Hence
$$
\lim_{x\to0} f(x)
=\lim_{x\to0} (f(x)-3e^{-x})+\lim_{x\to0} 3e^{-x}
=0+3
=3.
$$

