I got this doubt while solving this problem:
Given $g : \mathbb{R} \to \mathbb{R}$ $$g(x)=0$$
and $f: \mathbb{R} \to \mathbb{R}$
$$f(x)= \begin{cases} 0, \quad & \text{ if } x \ne 0, \\ 1, &\text{ if } x = 0\end{cases}$$
Now we have $$\lim_{x \to 0} f(x)=0 \ne f(0)$$ hence $f$ has Removable Discontinuity at $x=0$
Also $$\lim_{x \to 0} g(x)=0$$
Its quite obvious that $f(g(x)=1$ $\forall$ $x \in \mathbb{R}$
Hence $$\lim_{x \to 0} f\left(g(x)\right)=1$$
But $$f\left(\lim_{x \to 0}g(x)\right)=f(0)=1$$
So can Limit and function be interchanged when the function has Removable discontinuity?
No, in your example magically values are equal , so it does not mean you can take the limit inside.
Consider
$$g(x)=x^2$$
$$f(x)= \begin{cases} x+1, \quad & \text{ if } x \ne 1, \\ 4, &\text{ if } x = 1\end{cases}$$
we have $$f\left(g(x)\right)= \begin{cases} x^2+1, \quad & \text{ if } x \ne \pm1, \\ 4, &\text{ if } x = \pm1\end{cases}$$
So $$\lim_{x \to 1}f(g(x))=2$$ while
$$f\left(\lim_{x \to 1} g(x)\right)=4$$