Show that $f(x) = x\ln(x)$ is continuous on $(0,1)$ The easiest thing I can think of is to compute the derivative which comes out to be $ f'(x) = \ln(x) + 1 $. Since this is defined for all values on the interval $(0,1)$, the original function must be continuous.
However, I find $\epsilon - \delta$ proofs of continuity to be more... beautiful and elegant (and hard for me to get right). My attempt at that approach was this:
$$ |f(x) - f(k)| = |x\ln(x) - k\ln(k)| = |x\ln(x) - k\ln(x) + k\ln(x) - k\ln(k)| \\ \leq |x\ln(x) - k\ln(x)||k\ln(x) - k\ln(k)| \leq |x - k||\ln(x)||k||\ln(x) - \ln(k)| \\ \leq  |x - k||\ln(x)||\ln(x) - \ln(k)| $$
But I get stuck here. 
Is the derivative method okay? And what could I do to make the other method work? Thank you for your input!
 A: For the pure $\epsilon - \delta$ proof (without assuming a priori continuity of the log function) take $k \in (0,1)$ fixed.
Using the triangle inequality we have for all $x \in (0,1)$
$$\begin{align}|x \ln x - k \ln k| &= |x \ln x - x \ln k + x \ln k - k \ln k| \\ &\leqslant |\ln k||x -k| + |x||\ln x - \ln k| \\ &\leqslant |\ln k||x -k| + |\ln(x/k)|\end{align}$$
If $|x-k| < \delta_1 = \epsilon/(2|\ln k|)$ then $|\ln k||x- k| < \epsilon/2.$
For $x \geqslant k$ we have 
$$\left|\ln \frac{x}{k} \right| = \ln \frac{x}{k} = \ln \left( 1+ \frac{x-k}{k}\right) \leqslant \frac{x-k}{k} = \frac{|x-k|}{k}, $$
and if  $|x-k| < \delta_2 = k\epsilon/2$, then $|\ln(x/k)| < \epsilon/2$ when $x \geqslant k$.
Finally, for $x < k$ we have
$$\left|\ln \frac{x}{k} \right| = \ln \frac{k}{x} \leqslant \frac{|x-k|}{x}.$$
If $|x -k| < \delta_3 = \min(k/2, k\epsilon/4) $, then $x > k/2$ and 
$$\left|\ln \frac{x}{k} \right| \leqslant \frac{|x-k|}{x} < \frac{2|x-k|}{k} < \frac{\epsilon}{2}.$$
Putting it all together, if $|x - k| < \delta = \min(\delta_1,\delta_2, \delta_3)$, then $|x \ln x - k \ln k| < \epsilon$.
A: We just need to show that $g(x)=\ln x$ it's continuos on $(0,1)$, you can easily show $\displaystyle \lim_{x\to a}f(x)=\displaystyle \lim_{x\to a}x=a=f(a)$, then $f(x)$ it's continuous on $\mathbb{R}$, for $g(x)$, let $a>0$ we need to show $\displaystyle \lim_{x\to a}g(x)=g(a)$, in fact:
$$|g(x)-g(a)|=|\ln x-\ln a|=\left|\ln\left(\frac{x}{a}\right)\right|=\left|\ln\left(1+\frac{x-a}{a}\right)\right|\leq\frac{|x-a|}{a}$$
Then given $\epsilon>0$, exists $\delta(\epsilon)=\frac{\epsilon}{a}>0$ such that $0<|x-a|<\delta$ then $|g(x)-g(a)|<\epsilon$, i.e, $\displaystyle \lim_{x\to a}g(x)=g(a)$, now $h(x)=f(x)g(x)=x\ln x$ it's defined on $(0,\infty)$, since $f(x)$ and $g(x)$ are continuous functions then $h(x)$ it's continuos on $(0,\infty)$. Greetings.
