Find $\lim_{x \to 0} \frac{(1 + nx)^n - (1 + n^2x)}{x^2}$

Question

The title pretty much describes the problem. It must be achieved using elementary mathematical approaches (hence we must spare ourselves the pleasure of using L'Hopital).

What I've managed to do is the following rebuild of the function:

$$A = \frac{(1 + nx)^n - (1 + n^2x)}{x^2} =$$ $$= \frac{\sum_{k=0}^n\Big({n \choose k}n^kx^k\Big) - 1 - n^2x}{x^2} = $$ $$= \frac{1 + n^2x + \sum_{k=2}^n\Big({n \choose k}n^kx^k\Big) - 1 - n^2x}{x^2} =$$ $$= \frac{\sum_{k=2}^n\Big({n \choose k}n^kx^k\Big)}{x^2} =$$ $$= \sum_{k=2}^n{n \choose k}n^kx^{k-2}$$

Hence the problem is equivalent to finding: $$\lim_{x \to 0}\sum_{k=2}^n{n \choose k}n^kx^{k-2}$$

I suspect the answer is $\frac{n^3(n-1)}{2}$.

How would we go further? Thanks in advance!

Answer 1

Since $\displaystyle\sum_{k=2}^n\binom nkn^kx^{k-2}$ is a polynomial function, its limit at $0$ is its constant term, which is indeed $\frac{n^3(n-1)}2$.

Answer 2

You're nearly there! Now take limits. Is there any term without an $x$ in it?

Hint: $x^0 = 1$.