Project Euler 35 - Count time for 2 different algorithms

'''
The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.
There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
How many circular primes are there below one million?
'''
from math import ceil, pow

# First, I create a list with all the primes
LIMIT = list(int(pow(10, i)) for i in range(1, 5))
def f1(LIMIT):
    primes = [2]
    for i in range(3, LIMIT):
        isIPrime = True
        for prime in primes:
            if i % prime == 0:
                isIPrime = False
                break
        if isIPrime == True:
            primes.append(i)
    return primes

def f2(LIMIT):
    primes = [2]
    for i in range(3, LIMIT):
        isIPrime = True
        for j in range(2, ceil(i/2+1)):
            if i % j == 0:
                isIPrime = False
        if isIPrime == True:
            primes.append(i)
    return primes

print(LIMIT)
from timeit import default_timer as timer
times1 = []
times2 = []
for limit in LIMIT:
    start = timer()
    f1(limit)
    end = timer()
    times1.append((end - start) * 1000)

for limit in LIMIT:
    start = timer()
    f2(limit)
    end = timer()
    times2.append((end - start) * 1000)

print("Limit      1st Function                   2nd Function")
for i in range(len(LIMIT)):
    print(str(LIMIT[i]) + "         " + str(times1[i]) + "       " + str(times2[i]))