Difficult limit $\lim_{x\to\infty} e^{-x}\int_0^x \int_0^x \frac{e^u-e^v}{u-v} \,\mathrm{d}u\mathrm{d}v$ I need to calculate this limit:
$$\lim_{x\to\infty} e^{-x}\int_0^x \int_0^x \frac{e^u-e^v}{u-v} \,\mathrm{d}u\mathrm{d}v.$$
How do I do it?
There's a hint that I should use de l'Hospital's rule.

Expand the exponentials in their standard Taylor series to see the integrand equals
$$\sum_{n=0}^{\infty}\frac{1}{n!}\frac{u^n-v^n}{u-v}= \sum_{n=1}^{\infty}\frac{1}{n!}(u^{n-1} + u^{n-2}v + \cdots + uv^{n-2} + v^{n-1}).$$
For each $n$ we have
$$\int_0^x\int_0^x (u^{n-1} + u^{n-2}v + \cdots + uv^{n-2} + v^{n-1})\,dv\,du = x^{n+1}\left ( \frac{1}{n\cdot 1} + \frac{1}{(n-1)\cdot 2} + \frac{1}{(n-2)\cdot 3} +\cdots + \frac{1}{2\cdot (n-1)} + \frac{1}{1\cdot n}\right) \ge x^{n+1}\frac{1}{n}\left ( \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n}\right ) \ge x^{n+1}\frac{\ln n}{n}.$$
Thus the full expression of interest is at least
$$e^{-x}\sum_{n=1}^{\infty} \frac{\ln n}{n\cdot n!}x^{n+1} \ge e^{-x}\sum_{n=1}^{\infty} \ln n\frac{x^{n+1}}{(n+1)!}.$$
If we had just $e^{-x}\sum_{n=1}^{\infty} \frac{x^{n+1}}{(n+1)!}$ on the right, the limit on the right would be $1.$ But we have the extra $\ln n$ in front of the standard coefficients, hence the limit on the right is $\infty.$ Therefore the limit on the left is $\infty$ and we're done.
