calculus - Finding value that makes $f$ continuous

Let $$f(x)=\begin{cases}\dfrac{6}{x}+\dfrac{-5x+30}{x(x-5)}, & x \neq 0,5\\ \dfrac{1}{5}, & x= 0,5\end{cases}$$
Determine the value that makes $f$ continuous at $x=0$.

So we find the limit, and if the limit exists, then the limit is continuous, and the limit equals the functional value? I know how to mechanically solve this problem but I don't really understand what's going on and what I am doing. How are we redefining $f(0)$?

$\lim_{x \to 0} \dfrac{6x-30-5x+30}{x(x-5)}=\lim_{x \to 0} \dfrac{x}{x(x-5)}=-\dfrac{1}{5}$

$f(0)=-\dfrac{1}{5}$

But then what is the positive $\dfrac{1}{5}$ included in the problem? Why is the function continuous if it doesn't equal $\dfrac{1}{5}$?

How do you know which kind of discontinuities are at $x=0,5$? How do we know that at $0$ is removable discontinuity and at $5$ is infinite discontinuity?

Thank you.

Answer

The initial function is not continuous in $\ x=0$, because:

$$\lim_{x\to0}f(x)=-\frac{1}{5}≠\frac{1}{5}=f(0)$$

as you already proved; if instead it was defined in this way:

$$\ f(x) = \left\{
\begin{array}{l l}

\frac{6}{x}+\frac{-5x+30}{x(x-5)}, & \quad \text{$x≠0,5$}\\
-\frac{1}{5}, & \quad \text{$x=0,5$}
\end{array} \right.\ $$

then the function would be continuous in $\ x=0,$ but of course it keeps being not continuous for $\ x=5$, where there is an asymptote.