#2015_26 Learning Calculus!

$\begin{aligned} \lim_{n \to \infty} \frac{n^{1999}}{n^x - (n-1)^x} & = \lim_{n \to \infty} \frac{n^{1999}}{n^x - \left(n^x - xn^{x-1} + \frac{x(x-1)}{2}n^{x-2}-...\right)} \\ & = \lim_{n \to \infty} \frac{n^{1999}}{xn^{x-1} - \frac{x(x-1)}{2}n^{x-2} + \frac{x(x-1)(x-2)}{6}n^{x-3} - ...} \\ & = \lim_{n \to \infty} \frac{1}{xn^{x-2000} - \frac{x(x-1)}{2}n^{x-2001} + \frac{x(x-1)(x-2)}{6}n^{x-2002} ...} \quad \quad \color{#3D99F6}{\text{When } x = 2000} \\ & = \lim_{n \to \infty} \frac{1}{2000 - \frac{2000(1999)}{2n} + \frac{2000(1999)(1998)}{6n^2} ...} \\ & = \frac{1}{2000} \\ & \\ \Rightarrow x = \boxed{2000} \end{aligned}$

Since this is currently of the form (infinity/infinity), we may use LH rule of differentiation:

Limit $\frac{1999n^{1998}}{x(n^{x-1}-(n-1)^{x-1})}$

Let us assume x>1999. So we keep differentiating by LH rule till power of n in numerator become 0, so we get:

Limit $\frac{1999*1998*1997....2*1}{x(x-1)(x-2)...(x-1998)(n^{x-1999}-(n-1)^{x-1999})}$

Now since we still have a term of n in denominator we will get the value of limit =0 as n become infinite. But value of limit is 1/2000, so the power of n in denominator should also vanish, for which x=2000.

We may also check, in case x=2000, limit becomes

$\frac{1999*1998*1997....2*1}{2000*1999*1998...3*2(n-(n-1)}$

=

1999!/2000! = 1/2000

The same method can be used by assuming x<1999 and proceeding. In that case power of n will vanish first in denominator and then will have have to make power of n = 0 in numerator so that limit becomes finite. It will again yield x=2000