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$1)$ 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.

Did
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CAF
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  • (1) yes and (2) yes. The last part isn't actually required unless one wants to use a little more the definition of finite limit. – DonAntonio Mar 13 '13 at 18:11
  • Thanks DonAntonio. Why would one want to use little more than the defintion of a finite limit? – CAF Mar 13 '13 at 18:16
  • CAF, it may be that's the way limits are treated in general in that book, and since this is a beginner's exercise they want to make sure things are completely clear... – DonAntonio Mar 13 '13 at 18:31
  • The part where you state that "using the limit law $\lim\limits_{n \rightarrow \infty} \sqrt[n]{f(x)} = \sqrt[n]{\lim\limits_{n \rightarrow \infty} f(x)} $" is definitely not kosher. A problem is that any expression similar to $\sqrt[n]{\lim\limits_{n \rightarrow \infty} \ast} $ is meaningless. – Did Mar 21 '13 at 12:26
  • The second half we have seen before: http://math.stackexchange.com/questions/11/does-99999-1 – Ross Millikan Mar 21 '13 at 13:09
  • @Did: Could you elaborate? Isn't it just a definition that I wrote (since the square root is continuous on its domain) – CAF Mar 21 '13 at 19:23
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    One cannot consider the limit when n $\to\infty$ and then take the n th root of this limit. – Did Mar 21 '13 at 20:32

1 Answers1

-1

Another way for second question.

Let $x=0.999999999999\dots $

$10x=9.999999999999\dots$

subtract both equations

$9x=9$

$x=1;$

Ali Caglayan
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ABC
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