Why would this be a deterministic context free language? $L = \{a^n b^n | n>0 \}$ $\cup$ $\{ a^n b^{2n} | n < 100 \}$ so the book that I'm reading says this is a deterministic context free language 
 $L = \{a^n b^n | n&gt;0 \}$ $\cup$ $\{  a^n b^{2n} | n &lt; 100 \}$

But i think this is wrong Because :

at the beginning of building the DPDA, we have to decide whether we should go with the first one or the second, and considering both of them start with a there is just no way we could build a DPDA right?? or am i wrong? 

i also tried solving this by first building the $a^nb^n$ and then going for the second one if the stack is not empty, but still couldn't come up with anything, i tried putting 2 a's for every a input but didn't work

so how is the book saying this can be build with a DPDA?

Also i have to say I KNOW that non infinite languages are regular, but this one is not infinite!
Maybe you can create 100 states that push two $a$'s onto the stack for every $a$, and then pop these $a$'s for every $b$. In the event of getting to the 100th state, when you read the 101st $a$, pop 100 $b$s from the stack and continue as you would for the first language alone.