Jtdylktuy

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 $$int_0^1 x^ne^x dx=(-1)^{n+1}n!biggl(1-esum_{k=0}^nfrac{(-1)^k}{k!}biggr)$$ Now my question is this, how do I compute the limit of this expression as $n$ goes to $infty$? I recognize that $$lim_{ntoinfty}esum_{k=0}^{n}frac{(-1)^k}{k!}=1$$ 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,
begin{align*}
I_n&lemax_{xin[0,1]}{e^x}int_0^1x^{n}~mathrm dx\
&=frac{e}{n+1}to0.
end{align*}
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
$$
lim_{ntoinfty}int_0^y x^n e^x dx le
lim_{ntoinfty}y^nint_0^y e^x dx=0.
$$

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

$$exp(-1)=sum_{k=0}^nfrac{(-1)^k}{k!}+frac{e^xi}{(n+1)!}(-1)^{n+1}$$

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