mjhjmtu

I was doing some work on the integral $int_0^1 x^ne^x dx$ and I eventually came to this expression in terms of n $$int_0^1 x^ne^x dx=(-1)^n+1n!biggl(1-esum_k=0^nfrac(-1)^kk!biggr)$$ Now my question is this, how do I compute the limit of this expression as $n$ goes to $infty$? I recognize that $$lim_ntoinftyesum_k=0^nfrac(-1)^kk!=1$$ per the series expression of $e^x$ at $-1$. This gives me an indeterminate form, but I can't apply L'Hopital's rule because I can't differentiate $n!$ (or $1/n!$ as the case would be after re-arranging), and I also can't think of a variable substitution that would help me here. Common sense and wolfram definitely imply that the limit is in fact $0$, but how can I show this?

When $xin [0,1]$, $e^xle e$, by first mean value theorem for definite integrals,
beginalign*
I_n&lemax_xin[0,1]e^xint_0^1x^n~mathrm dx\
&=fracen+1to0.
endalign*
Since $I_n>0$, the limit is $0$ by squeeze theorem.

Not as slick as Tianlalu's method, but for $yin (0,,1)$ we have
$$
lim_ntoinftyint_0^y x^n e^x dx le
lim_ntoinftyy^nint_0^y e^x dx=0.
$$

I don't know what you are 'allowed' to use but you could use the lagrange remainer for the exponential function:

$$exp(-1)=sum_k=0^nfrac(-1)^kk!+frace^xi(n+1)!(-1)^n+1$$

for some $-1<xi<0$. If you estimate $e^xi $ and replace the sum you get convergence to $0$.