Limits $\sqrt{2n^2-1}/(n+1)$ and $1 = 0.9999...$ $1)$  Find the limit (if it exists) of the following sequence:
$$\frac{\sqrt{2n^2-1}}{n+1} = x_n$$

Attempt: Rewrite as $$\frac{\sqrt{n^2(2 - \frac{1}{n^2})}}{n+1} = \frac{n\sqrt{(2-\frac{1}{n^2})}}{n+1} = \frac{\sqrt{2 - \frac{1}{n^2}}}{\frac{1}{n} + 1}$$ So as $n \rightarrow \infty,\,\,x_n \rightarrow \sqrt{2},$ using the limit law $\operatorname{lim}_{n \rightarrow \infty} \sqrt[n]{f(x)} = \sqrt[n]{\operatorname{lim}_{n \rightarrow \infty} f(x)} $and clearly $f(x) \geq 0 \,$for $n \neq 0$ 

Is the above calculation rigorous enough to determine the limit?

2)Interpret a decimal expansion $0.a_1a_2a_3....$ as $$0.a_1a_2a_3.... = \operatorname{lim}_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{a_k}{10^k}.$$
Show that $1 = 0.9999...$

Attempt: $$0.9999... =\operatorname{lim}_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{a_k}{10^k} = 9 \cdot \sum_{k=1}^{\infty} \frac{1}{10^k} = 9 \cdot \frac{1/10}{9/10} = 9 \cdot \frac{1}{9} = 1$$ Done?

In my book, they also include the following:Let $ y_n = \sum_{k=1}^{\infty} \frac{9}{10^k}$ and we want $|1-y_n| = |1 -  \sum_{k=1}^{\infty} \frac{9}{10^k}|$ As shown above the RHS tends to 0 so $|1-y_n| \rightarrow 0 $as $n \rightarrow \infty$ 
What does this add and is it required?

Many thanks.
Another way for second question.

Let $x=0.999999999999\dots $

$10x=9.999999999999\dots$

subtract  both equations

$9x=9$

$x=1;$
Another way for second question.

Let $x=0.999999999999\dots $

$10x=9.999999999999\dots$

subtract  both equations

$9x=9$

$x=1;$