Show $\lim\limits_{x\to0}x\ln(x)=0$ by using Taylor series Is it possible to show the limit $\lim\limits_{x\to0}x\ln(x)=0$ by expanding the corresponding Taylor series?

My approach:
I know that the Taylor series of $\ln(x)$ converges uniformly on a closed interval $(1-\delta,1+\delta)$ where $\delta>0$ is sufficiently small.
So let be $$S_n(x):=\sum\limits_{k=1}^n(-1)^{k+1}(x-1)^k\frac{1}{k}\\
\lim\limits_{n\to\infty}S_n(x)=\sum\limits_{k=1}^{\infty}(-1)^{k+1}(x-1)^k\frac{1}{k}=\ln(x)\\
T_n(x):=(x-1)+\sum\limits_{k=2}^n(-1)^{k}(x-1)^k\frac{1}{(k-1)k}\\
\lim\limits_{n\to\infty}T_n(x)=(x-1)+\sum\limits_{k=2}^{\infty}(-1)^{k}(x-1)^k\frac{1}{(k-1)k}\overset{?}{=}x\ln(x),
$$
where $\lim\limits_{n\to\infty}S_n(x)$ denotes the Taylor series of $\ln(x)$ at $x=1$ and $\lim\limits_{n\to\infty}T_n(x)$ the Taylor series of $x\ln(x)$ at $x=1$.
What I can observe so far is that $T_n(x)$ converges uniformly on $(0,1+\delta)$ (the point $x=0$ doesn't cause problems anymore) and this allows us to swap limit taking:
$$
\lim\limits_{x\to0}\lim\limits_{n\to\infty}T_n(x)=\lim\limits_{n\to\infty}\lim\limits_{x\to0}T_n(x)=0.
$$
However, I don't know what the limit function is or in other words how to show that $\lim\limits_{n\to\infty}T_n(x)=x\ln(x)$.

Any suggestions how to complete the proof?
 A: It helps to keep in mind the geometric series $$\sum_{k=0}^\infty (1-x)^k = \frac 1{1-(1-x)} = \frac 1x$$
which is valid for $x \in (0,2)$.  You have that the Taylor expansion $$\ln x = \sum_{k=1}^\infty \frac{(-1)^{k+1}}k(x-1)^k$$ is valid on the interval $(0,2)$ but is not uniform on any interval of the form $(0,\epsilon)$. However you can conclude that $x \ln x$ is bounded on $(0,1)$ because if $0 < x < 1$ then
$$|x \ln x| = \left| \sum_{k=1}^\infty \frac{(-1)^{k+1}}kx(x-1)^k \right| \le \sum_{k=0}^\infty x(1-x)^k = 1.$$
You can finish up rather quickly if you are willing to use a law of logarithms. If $0 < x < 1$ then $0 < x^{1/2} < 1$ too and $$|x^{1/2} \ln x| = 2 |x^{1/2} \ln x^{1/2}| \le 2$$ so that $$|x \ln x| \le 2|x|^{1/2}.$$
A: I'm surprised this question still didn't produce the answer the OP is missing (badly). First off: the Taylor series of $x\,\ln x$ around $x=1$ is indeed (correcting a typo from above) $$x-1+\sum^\infty_{k=2}\frac{(-1)^k}{k\,(k-1)}\,(x-1)^k.\tag{1}$$ That can be done (as the OP did) by multiplying the known series for $\ln x$ by $x=1+(x-1)$, or directly by calculating the derivatives at $x=1$, observing $$\frac{d}{dx}\,x\,\ln x=\ln x+1.$$ Whether (1) really converges to $x\,\ln x$, is another question, but if we assume (as the OP obviously did) that the series for $\ln x$ converges for $x\in(0,2)$, their derivation shows that (1) converges in that interval, too. Otherwise, we might just estimate the remainder term to come to the same conclusion.
Concerning the limit as $x\to0$, we have Abel's theorem (https://en.wikipedia.org/wiki/Abel%27s_theorem), saying that $$\lim_{x\to0} x\,\ln x=-1+\sum^\infty_{k=2}\frac1{k\,(k-1)}=-1+\sum^\infty_{k=2}\left(\frac1{k-1}-\frac1k\right)=-1+1=0,$$ since the series on the RHS converges.
