Is function $f(x) = \frac{x^{2}-2}{x-\sqrt{2}}$ continuous for all $x$? 
Is function $f(x) =  \frac{x^{2}-2}{x-\sqrt{2}}$ continuous for all $x$?

I understand that the function
$$ g(x) = \frac{x^{2}-1}{x-1} $$
has one discontinuity for $x=1$. 

[Added later] In general my question is for the function 
$$
\frac{x^{2}-a}{x-\sqrt a},
$$
Can we say that there is a discontinuity at point $x = a$? If $a$ is an irrational number, then how to define its point of discontinuity?
 A: you can write $$f(x)=\frac{(x-\sqrt{2})(x+\sqrt{2})}{x-\sqrt{2}}=x+\sqrt{2}$$ for $$x\neq \sqrt{2}$$ and if you define $$f(x)=x+\sqrt{2}$$ for $$x\neq \sqrt{2}$$ and $$f(x)=2\sqrt{2}$$ if $x=\sqrt{2}$ then your function is continuous
A: $\require{cancel}$
Indeed, $$f(x) =  \frac{x^{2}-2}{x-\sqrt{2}}$$ has a discontinuity at $x=\sqrt 2$ (the point at which the domominator is 0).  But it is removable.
$$f(x) = \frac{x^2 - 2}{x-\sqrt 2} =\frac{x^2 - (\sqrt 2)^2}{x-\sqrt 2}= \frac{(x+\sqrt 2)(\cancel{x-\sqrt 2})}{\cancel{x-\sqrt 2}} = {x+\sqrt 2}$$ when $x \neq \sqrt 2$.  
All that remains is to define $f(\sqrt 2) = 2\sqrt 2$.  Now, we have defined the function so that $f(x)$ is continuous!
$$f(x) = \begin{cases} x+\sqrt 2 & x \neq \sqrt 2\\ \\2\sqrt 2 & x = \sqrt 2\end{cases}$$

In general, suppose $c\in \mathbb R$. Then, the function defined by $$f(x) = \frac{x^2−c^2}{x−c}$$ is continuous everywhere in $\mathbb R$, except at the point $x=c$. 
Such a discontinuity is removable by defining $f(x) = x+c$, for all $x\neq c$, and otherwise,  $f(c)=2c$
