Show continuity of $\sqrt{1-x} \sum_{k=1}^\infty \frac{x^k}{\sqrt{k}}$ Show that the function $f:[-1,1]\to \mathbb{R}$
$$
f(x):=\sqrt{1-x} \sum_{k=1}^\infty  \frac{x^k}{\sqrt{k}}
$$
is well-defined and continuous at the point $1$.
 A: 
Suppose that $a(x)=\sum\limits_{n=1}^{\infty}a_n x^n$ and $b(x)=\sum\limits_{n=1}^{\infty}b_n x^n$ converge when $|x|<1$, all $b_n$ are positive, and $\lim\limits_{x\ \uparrow\ 1}b(x)=\infty$. If $\lim\limits_{n\to\infty}\dfrac{a_n}{b_n}=\lambda$ exists then $\lim\limits_{x\ \uparrow\ 1}\dfrac{a(x)}{b(x)}=\lambda$.

For a proof, replacing $a(x)$ with $a(x)-\lambda b(x)$, we may assume that $\lambda=0$. Now if $\varepsilon>0$ is arbitrary, and $|a_n/b_n|<\varepsilon$ when $n>N$, then for $0<x<1$
$$\left|\frac{a(x)}{b(x)}\right|\leqslant\frac{1}{b(x)}\left|\sum_{n=1}^{N}a_n x^n\right|+\varepsilon\underset{x\ \uparrow\ 1}{\longrightarrow}\varepsilon,$$
i.e. $\limsup\limits_{x\ \uparrow\ 1}|a(x)/b(x)|\leqslant\varepsilon$. As $\varepsilon$ is arbitrary, we have $\lim\limits_{x\ \uparrow\ 1}\big(a(x)/b(x)\big)=0$ as claimed.
Now apply the above to
$$a(x)=\sum_{n=1}^{\infty}\frac{x^n}{\sqrt{n}},\quad b(x)=\frac{1}{\sqrt{1-x}}-1=\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!}x^n.$$
We have $\displaystyle\lim_{n\to\infty}\frac{(2n)!!}{(2n-1)!!\sqrt{n}}=\sqrt{\pi}$ by Wallis formula, thus $\lim\limits_{x\ \uparrow\ 1}f(x)=\sqrt{\pi}$ as well.
[For the "previous version" we get $\lim\limits_{x\to\pi/2}f(x)\cos x=1$ the same way.]
