Use the ratio test: Let an = n!/en

lim n→∞ |an+1/an|

= lim n→∞ |[(n + 1)!/en+1]/(n!/en)

= lim n→∞ |[(n + 1)!/en+1](en/n!)

= lim n→∞ |[(n + 1)n!en]/(enn!e)

= (1/e) lim n→∞ (n + 1) = ∞, so the given series diverges.

f(x) = x/2

Then the differential is lim h->0 [f(x+h) - f(x)]/h

= lim h->0 [(x+h)/2 - x/2]/h

= lim h->0 [h/2]/h

= lim h->0 [1/2] = 1/2

f(x) = x/2

Then the differential is lim h->0 [f(x+h) - f(x)]/h

= lim h->0 [(x+h)/2 - x/2]/h

= lim h->0 [h/2]/h

= lim h->0 [1/2] = 1/2

f(x) = x/2

Then the differential is lim h->0 [f(x+h) - f(x)]/h

= lim h->0 [(x+h)/2 - x/2]/h

= lim h->0 [h/2]/h

= lim h->0 [1/2] = 1/2

f(x) = x/2

Then the differential is lim h->0 [f(x+h) - f(x)]/h

= lim h->0 [(x+h)/2 - x/2]/h

= lim h->0 [h/2]/h

= lim h->0 [1/2] = 1/2

This browser is totally bloody useless for mathematical display but...

The probability function of the binomial distribution is P(X = r) = (nCr)*p^r*(1-p)^(n-r) where nCr =n!/[r!(n-r)!]

Let n -> infinity while np = L, a constant, so that p = L/n

then

P(X = r) = lim as n -> infinity of n*(n-1)*...*(n-k+1)/r! * (L/n)^r * (1 - L/n)^(n-r)

= lim as n -> infinity of {n^r - O[(n)^(k-1)]}/r! * (L^r/n^r) * (1 - L/n)^(n-r)

= lim as n -> infinity of 1/r! * (L^r) * (1 - L/n)^(n-r) (cancelling out n^r and removing O(n)^(r-1) as being insignificantly smaller than the denominator, n^r)

= lim as n -> infinity of (L^r) / r! * (1 - L/n)^(n-r)

Now lim n -> infinity of (1 - L/n)^n = e^(-L)

and lim n -> infinity of (1 - L/n)^r = lim (1 - 0)^r = 1

lim as n -> infinity of (1 - L/n)^(n-r) = e^(-L)

So P(X = r) = L^r * e^(-L)/r! which is the probability function of the Poisson distribution with parameter L.