question about a simple limit of $\int_{e^{-1}}^{1}x^{n}\cdot \ln(x)\,dx$ limit of $\int_{e^{-1}}^{1}x^{n}\cdot \ln(x)dx$ 
I had this integral at exam and I have some questions about it. The first method is to solve the integral then to find the limit.
I solved it like this:
$e^{-1}\leq x\leq 1\Rightarrow -1\leq \ln(x)\leq 0$ 
$ -x^{n}\leq \ln(x)\cdot x^{n}\leq 0$. I applied the integral, I solved it and by squeezing theorem, thus the limit is $0$.
Is my solution right?My friends told me that it's not correct because I didn't changed the signs like $-1\geq  \ln(x)\geq  0$ but ln function is an incrasing function and the base is $e$ not $(0,1)$.
Should I change signs?
 A: You can integrate by parts, and there is no need to make a substitution in the limits. You end up with:
$\frac{e^{-(n+1)}(n+2)-1}{(n+1)^2}$
Then, assuming you're finding the limit as $x \rightarrow \infty$, we can see that the $e^{-(n+1)}$ and the $(n+1)^2$ "outweigh" the linear term in the numerator and cause the denominator to increase at a faster rate. Hence, the expression tends to 0 (from below).
A: $$
\begin{align}
\int_{e^{-1}}^1x^n\log(x)\,\mathrm{d}x
&=\frac1{(n+1)^2}\int_{e^{-n-1}}^1\log(x)\,\mathrm{d}x\tag1\\
&=\frac1{(n+1)^2}\left[x\log(x)-x\vphantom{\int}\right]_{e^{-n-1}}^1\tag2\\[3pt]
&=\frac{-1+(n+2)e^{-n-1}}{(n+1)^2}\tag3
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto x^{\frac1{n+1}}$
$(2)$: integrate
$(3)$: evaluate
A: Hint: One method may be interested is the inequality 
$$\dfrac{x-1}{x}<\ln x<x-1$$
hold $x>0$, then both sides are the form 
$$\int_{\frac1e}^{1}x^{m+1}-x^{m}\ dx$$
which tends to zero as $n\to\infty$.
A: You are correct. Another approach is to simply blast through with absolute values:
$$|\int_{1/e}^1 x^n\ln x\,dx|\le \int_{1/e}^1 |x^n\ln x|\,dx =\int_{1/e}^1 x^n|\ln x|\,dx.$$
Since $|\ln x|$ is bounded by some $M$ on the interval of integration, we can bound the last integral above by
$$M\int_{1/e}^1 x^n\,dx < M\int_{0}^1 x^n\,dx = \frac{M}{n+1} \to 0.$$
