Continuity, removable and essential discontinuity I want to know if the following 2 functions are continuous or not.  
1.
$$
f(x) =
\begin{cases}
1/x{^2}  & \text{if } x \ne 0 \\
2 & \text{ if } x=0  \\
\end{cases}.
$$
 Then I want to check the continuity at $0$,
 $$
\lim_{x\to 0}f(x)=\infty, \quad f(0)=2.
$$
Since $f(0)$ is not equal to $\lim_{x\to 0}f(x)$,  $f$ is not continuous at $0$.
Since $\lim_{x\to 0}f(x)$ exists (going to infinity doesn't mean that limit doesn't exist right?) and $f$ is not continuous at $0$, isn't this a removable discontinuity?
Because if limit at that point exists and f is not continuous at that point it is a removable discontinuity and otherwise it is essential discontinuity.  
2) 
$$
f(x) =
\begin{cases}
x^2  & \text{if } x \ne 2 \\
4 & \text{ if } x=2  \\
\end{cases}.
$$
Clearly this is continuous at $x=2$. Isn't it? But the lecturer said this is a removable discontinuity. Isn't he making a mistake here?
 A: 
So does a limit approaches infinity and the term limit does not exist are the same. 

Limits do not approach anything. Functions do. When a function approaches something, we say that its limit is equal to that thing. 
Anyway, the terms are not equivalent. The limit $\lim_{x\to 0} \sin (1/x)$ does not exist, but it would not be correct to say that this function approaches infinity; it clearly does not. 
However, it is true that when a function approaches infinity, the limit (understood in the strict sense of a limit in $\mathbb R$) does not exist. The failure of limit to exist in this particular way is recorded as "$\lim_{x\to a}f(x)=\infty$", a shorthand for "limit does not exist due to $f$ approaching infinity".  
Yet, there are ways in which one can extend the definition of  limit to make $\lim_{x\to a}f(x)=\infty$ an actual   limit rather than a shorthand form of the statement quoted above. This is a matter of how a particular book / professor define things.

Concerning the second part: you are right that $$f(x) =
\begin{cases}
x^2  & \text{if } x \ne 2 \\
4 & \text{if } x=2  \\
\end{cases}$$
is a continuous function. It is the same function as "$f(x)=x^2$ for all $x$", only written differently. 
Perhaps  what the lecturer had in mind was that 
$$f(x) =
\begin{cases}
x^2  & \text{if } x \ne 2 \\
\text{undefined} & \text{if } x=2  \\
\end{cases}$$
has a removable discontinuity, which is demonstrated by the fact that defining $f(2)=4$ makes a continuous function.
