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?

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).
