Consider the following grammar:

$S\rightarrow 0A\mid 0BB$

$A\rightarrow 00A\mid \lambda$

$B\rightarrow 1B\mid 11C$

$C\rightarrow B$

Which language does this grammar generate?

From production:

$S\rightarrow 0A$

and

$A\rightarrow 00A\mid \lambda$

So, $A$ can generate:

$\lambda,\ 00,\ 0000,\ldots$

Hence, $S$ generates:

$0,\ 000,\ 00000,\ldots$

This is represented by:

$(00)^*0$

Now consider:

$S\rightarrow 0BB$

But $B$ has productions:

$B\rightarrow 1B\mid 11C$

and

$C\rightarrow B$

Here, $B$ never terminates into $\lambda$ or any final terminal-only string.

So, this branch does not generate any valid finite string.

Hence, the grammar generates only:

$L((00)^*0)$