Limit revisited...

Let the limit be some value L. Take logarithm on both sides,

Lim_{n→∞} [nlog(x) - log(n!)] = logL
This is the natural logarithm. For log(n!), we use Stirling's formula -:
log(n!) = nlog(n) - n + more terms which can be neglected.

Lim_{n→∞} [nlog(x) - nlog(n) + n] = logL

Now you can see that the LHS will always be infinity whatever be the value of x.
logL = ∞, this happens when L = 0.
Hence the limit is 0.

Maybe his higher level concept involved gamma functions...

@ pritish ur sayin log0 =∞
but then 0= e^∞= ∞......???

There are two ways to look at it

1) By the archimedean property of real numbers there exists a natural number N such that N>x (i took the liberty of using this concept as u said u r in engg. otherwise you say its just plain common sense :D)

Let K=\frac{x^N}{N!}

and let L = \frac{x}{N+1}<1

Then

0< \frac{x^n}{n!} < K L^{n-N} = \left(\frac{K}{L^N}\right) L^n

and since \left(\frac{K}{L^N}\right)\lim_{n \rightarrow \infty} L^n = 0 by sandwich theorem the required limit is zero.

2) We know that if the infinite series tex]\sum_{n=1}^{\infty}a_n is convergent then \lim_{n\rightarrow \infty}a_n=0

This follows from the Cauchy criterion on the series a_1, a_1+a_2, a_1+a_2+a_3,..., a_1+a_2+...+a_n,...

Since e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} is convergent we must have \lim_{n \rightarrow \infty} \frac{x^n}{n!} = 0

sir i did not understand how this came..

0< \frac{x^n}{n!} < K L^{n-N} = \left(\frac{K}{L^N}\right) L^n....

sir i cud'nt understand how are we getting xⁿn! < K L^n-N ???

\frac{x^n}{n!} = \frac{x^N}{N!} \times \frac{x}{N+1} \times \frac{x}{N+2}\times...\times \frac{x^n}{n}

Now its obvious that \frac{x}{N+k} < \frac{x}{N+1} for k>1.

Hence

RHS<K \left(\frac{x}{N+1}\right)^{n-N} = KL^{n-N}

There was bound to be something wrong with my solution :P

Yes now I understood it...Thank you sir...