determine for which values of $x$ the function $f(x)=\begin{cases} \frac{x^2-4}{x-2}+10, &\text{if }x < 2\\ 2x^3-x, &\text{if } x ≥ 2\\ \end{cases}$ is continuous...
$$f(x)=\begin{cases}
\frac{x^2-4}{x-2}+10, &\text{if }x < 2\\
2x^3-x, &\text{if } x ≥ 2\\
\end{cases}$$
These kind of problems are very unclear to me, I only know these kind of functions have several graphs, in this example There are 2, But I can't figure out what these intervals means and even worse I don't even know where to start from.
 A: As Zubin Mukerjee indicated in the comments, we must show that 


*

*$f$ is continuous when $x < 2$;

*$f$ is continuous when $x > 2$;

*$\lim_{x \to 2+} f(x) = \lim_{x \to 2-} f(x) = f(2)$.  


If $x < 2$, then
\begin{align*} 
f(x) & = \frac{x^2 - 4}{x - 2} + 10\\
     & = \frac{(x + 2)(x - 2)}{x - 2} + 10\\
     & = x + 2 + 10 && \text{provided $x \neq 2$}\\
     & = x + 12
\end{align*}
Since $x < 2$, we are not dividing by zero in the second step.  Hence, the given expression is equivalent to $x + 12$ in the interval $(-\infty, 2)$.  Since $x + 12$ is a polynomial, $f$ is continuous in the interval $(-\infty, 2)$.
If $x > 2$, then $f(x) = 2x^3 - x$.  Since this is a polynomial, $f$ is continuous the interval $(2, \infty)$.  
Since 
$$\lim_{x \to 2+} f(x) = \lim_{x \to 2+} (2x^3 - x) = 2 \cdot 2^3 - 2 = 16 - 2 = 14$$ 
and 
$$\lim_{x \to 2-} f(x) = \lim_{x \to 2-} (x + 12) = 2 + 12 = 14$$
and 
$$f(2) = 2 \cdot 2^3 - 2 = 16 - 12 = 14$$
it is true that 
$$\lim_{x \to 2+} f(x) = \lim_{x \to 2-} f(x) = f(2)$$
Since we have verified all three conditions, $f$ is continuous on $(-\infty, \infty)$.  
A: This would be a piecewise function! It is broken into pieces. Now let's start.
$$f(x)=\begin{cases} \frac{x^2-4}{x-2}+10, &\text{if }x < 2\\ 2x^3-x, &\text{if } x ≥ 2\\ \end{cases}$$
For a function to be continuous on a point we must have
$$\lim_{x \rightarrow a}f(x)=f(a)$$
Limit of a function exists only when 
$$\lim_{x \rightarrow a^+}f(x)=\lim_{x \rightarrow a^-}f(x)$$
If $\lim_{x \rightarrow a^+}f(x)\neq \lim_{x \rightarrow a^-}f(x)$
then we say the limit does not exist!
Deal with left hand limit(values of x that is smaller than a) first
$$\lim_{x \rightarrow 2^-}\frac{x^2-4}{x-2}+10$$
Notice that direction substitution will result in indeterminate form $\frac{0}{0}$
The function can be redefined in a way such that
$$\lim_{x \rightarrow 2^-}\frac{x^2-4}{x-2}+10=\lim_{x \rightarrow 2^-}\frac{(x-2)(x+2)}{x-2}+10$$
$$\lim_{x \rightarrow 2^-}\frac{(x-2)(x+2)}{x-2}+10=\lim_{x \rightarrow 2^-}(x+2)+10$$
$$\lim_{x \rightarrow 2^-}(x+2)+10=14$$
Right hand limit
$$ \lim_{x \rightarrow 2^+}2x^3-x=2(2)^3-2$$
$$2(2)^3-2=14$$
Both sides limit are equal so we claim the limit exists and is equal to $14$
Now how about point $x=2$ we choose the following function because it contains the point that we want
$$f(x)=2x^3-x$$
$$f(2)=2(2)^3-2$$
$$f(2)=14$$
So we say that the function is continuous at $x=2$ by
$$\lim_{x \rightarrow 2}f(x)=f(2)$$
Some visual confirmation!

Notice that as x gets closer and closer to $2$ from the right side for the function $\frac{x^2-4}{x-2}+10$. Example, $1,1.5,1.66,1.99$
The function is approaching the value of $14$
In limit we don't care about what happen in at point $x=a$ We only care about what happen around it. In this case the right hand limit and the left hand limit are equal. Ergo, the limit exists.
If we were to graph the function 
$$\frac{x^2-4}{x-2}+10$$
alone  for domain $(- \infty,2)(2,\infty)$
A gap will happen at point $x=2$ The gap is removable hence called removable discontinuity! 
