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?

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}
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 $$\lim_{x\to 0^+} x\log x = \lim_{u\to\infty} \frac1u\log\big(\frac1u\big) = \lim_{u\to\infty}\frac{-\log(u)}u = 0.$$