Tangent space of quasi-projective varieties If $X$ is a quasi-projective variety and $X_i,\;\;i=1,\ldots,k\;$ are its irreducible components, then why $$\mathrm{dim}\;T_{X,x}=\mathrm{max}_{i=1,\ldots,k}\;(\mathrm{dim}\;T_{X_i,x})?\qquad \qquad (x\in X)$$ 
 A: What definition of $T_{X,x}$ are you using?  With at least one reasonable interpretation (i.e. the Zariski tangent space), the statement you give is false.
E.g. if $X = \{(x,y) \in \mathbb A^2 \, | \, xy = 0\}$ (two lines crossing at the origin) then the tangent space of the origin in each component is one-dimensional (since each component is smooth, being a line), but the Zariski tangent to the pair of lines at the origin is two-dimensional.
Added in response to the discussion in comments below:
Let $Z$ be an irreducible component of $X$ passing through $x$.  Then
$\dim T_{x,Z} \leq \dim T_{x,X}$.  
The proof is as follows: The local ring $\mathcal O_{x,Z}$ is a quotient of the
local ring $\mathcal O_{x,X}$, by an ideal $I$, say.  Then $\mathfrak m_{x,Z} = \mathfrak m_{x,Z}/I,$ and so $\mathfrak m_{x,Z}/\mathfrak m_{x,Z}^2 =
\mathfrak m_{x,X}/(I + \mathfrak m_{x,X})^2$ is a quotient
of $\mathfrak m_{x,X}/\mathfrak m_{x,X}^2$.  Passing to duals, we find
that $T_{x,Z}$ embeds as a subspace of $T_{x,X}$, which gives the required bound.
Taking the maximum over all components $Z$ passing through $x$, we find
that 
$$\max_{Z \text{ a comp. of $X$ passing through $z$}} \dim T_{x,Z} 
\leq \dim T_{x,Z}.$$
As the example given above shows, this inequality can sometimes be strict.
