..+x^2+x+1)}{x-1}$
Notice that we can cancel out the $(x-1)$ terms in the numerator and denominator, since $x-1$ is not equal to zero when $x$ approaches 1.
So, we are left with:
$\lim _{x \rightarrow 1} (x^{999}+x^{998}+...+x^2+x+1)$
Now, let's substitute $x=1$
Show more…