For $T(n) = 3T(n/3) + (n/2)$
with $a=3, b=3$ and $\log_3 3 = 1$
It's clear that $f(n) = \frac{1}{2}n$ is asymptoticaly slower than $f(n) = n^{\log_b a-e}$ for $e>0$ but the answer says it's case $2$.
where i am wrong to solve this ?
For $T(n) = 3T(n/3) + (n/2)$
with $a=3, b=3$ and $\log_3 3 = 1$
It's clear that $f(n) = \frac{1}{2}n$ is asymptoticaly slower than $f(n) = n^{\log_b a-e}$ for $e>0$ but the answer says it's case $2$.
where i am wrong to solve this ?
It is not asymptotically slower. Both are $\Theta(n)$ - they just have a different constant.
Note that $c_{crit} = 1$.