Why is $\int_{0}^{1} \log x $ computable? While evaluating $$\lim_{n \to \infty}\left(\frac{n!}{n}\right)^{1/n} $$ 
The integral turns out to be
$$\int_0^1 \log x = \big[x\log x\big]_0^1 -  \big[x \big]_0^1 $$ 
The second term will -1 how ever the first term will be
$$I_1=1\times\log 1 -0\times \log 0$$ 
My text book has taken the value of $0\log 0$ as $0$. However isn't $\log0$ undefined?
 A: Since $\log(0)$ is undefined, you have to take the one-sided (!) limit:
$$
L
:= \lim_{x \searrow 0} x \cdot \log(x)
= 0.
$$
This is zero, which you can obtain by using L'Hospitals rule:
\begin{align}
L
= \lim_{x \searrow 0} \frac{\log(x)}{\frac{1}{x}}
\overset{\text{L'H}}{=} \lim_{x \searrow 0} \frac{\frac{1}{x}}{- \frac{1}{x^2}}
= \lim_{x \searrow 0} - \frac{x^2}{x}
= \lim_{x \searrow 0} -x
= 0.
\end{align}

Basically, since log(x) isn't defined for one of the bounds of integration, I would always begin by writing
\begin{align}
\int_{0}^{1} \log(x) dx
& = \lim_{a \searrow 0} \int_{a}^{1} \log(x) dx
= \lim_{a \searrow 0} \big[ x \left( \log(x) - 1\right) \big]_{a}^{1} \\
& = \lim_{a \searrow 0} 1 ( 0 - 1) - a \left( \log(a) - 1\right)
=  -1 - \lim_{x \searrow 0} x \left( \log(x) - 1\right)
\end{align}
A: Rather than (ab)using L'Hôpital's rule, you should know (perhaps once and for all by L'Hôpital, but there are other ways) that as $x\to\infty$ exponential functions beat out any power of $x$. It follows that as $x\to\infty$ any power of the logarithm loses to $x$, i.e.,
$$\lim_{x\to\infty} \frac{(\log x)^n}x = 0.$$
Now, note that
$$\lim_{x\to 0^+} x\log x = \lim_{u\to\infty} \frac1u\log\big(\frac1u\big) = \lim_{u\to\infty}\frac{-\log(u)}u = 0.$$
