Limit involving $(1+x)^x$ term

Question

I don't know how to solve the following limit without using series expansion.

$$\lim_{x\to 0} \frac{(1+x)^x -1 -x^2}{x^3} $$

I have tried use L'Hopital's rule and finding bounds to use squeeze theorem but to no avail. Please give me hints on how to compute the limit, instead of posting full answers. Thanks in advance :)

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Edit: I have just found an answer, though it is not elegant at all. It is done simply by applying L'Hopital's rule three times and there is not much to say about it. Still, other answers are welcome.

Answer 1

Consider the limit with $x=1/p$, $p$ integer, $$\lim_{p\to\infty}p^3\left(\left(1+\frac1p\right)^{1/p}-1-\frac1{p^2}\right).$$

Multiplying/dividing by the conjugate multinomial we get

$$\lim_{n\to\infty}\frac{p^3\left(1+\frac1p-\left(1+\frac1{p^2}\right)^p\right) } {\left(1+\frac1p\right)^{(p-1)/p}+\left(1+\frac1p\right)^{(p-2)/p}\left(1+\frac1{p^2}\right)+\cdots\left(1+\frac1{p^2}\right)^{(p-1)/p}}.$$

By the binomial theorem, the numerator is the finite sum

$$p^3\left(1+\frac1p-1-\frac p{p^2}-\frac{p(p-1)}{2p^4}-\frac{p(p-1)(p-2)}{3!p^6}-\cdots\frac1{p^{2p}}\right)\\ =-\frac{(p-1)}{2}-\frac{(p-1)(p-2)}{3!p^2}-\cdots\frac1{p^{2p-3}}$$ while the denominator has $p$ terms that are all between $1$ and $1+1/p$, hence is between $p$ and $p+1$.

In the limit, a single term remains, $$-\frac12.$$

Answer 2

Hint:

You can use the generalized binomial formula, which works for any exponent.

$$(1+x)^x\approx 1+x\cdot x+\frac{x(x-1)}2x^2+\frac{x(x-1)(x-2)}{3!}x^3+\cdots\binom xn{x^n}+\cdots$$

The coefficient of $x^3$ is $-\dfrac12$.