Suppose that $X_1,\ldots,X_n$ is an iid sample from the Poisson distribution with mean $\lambda$. Use the Central Limit theorem to find $P(|\bar X - \lambda| < 0.1) $ as $n$ goes to infinity.
My question is, if $n$ goes to infinity, the variance for $\bar X$ would be zero and it does not make any sense to me.
For finite n, we have
$var(\bar X) > 0 $,
but $var(\bar X) \rightarrow 0$ for $n \rightarrow \infty$
Since $\bar X \rightarrow \lambda\ ,$
the probability you mentioned tends to 1, because for some n,
we have $P(|\bar X-\lambda|<0.1) > 1-\epsilon$
for each given $\epsilon>0$
Or, shortly formulated :
$$\lim_{n \rightarrow \infty} P(|\bar X-\lambda|<0.1)=1$$
This type of convergence is called convergence in probability.
For large $n$, $\bar X$ is nearly normal-distributed and since the variance tends to 0, $\bar X$ tends to the constant random-variable having the value $\lambda$ with probability $1$.
The $0.1$ can, by the way, be replaced by any positive value.