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1) Let $\{a_n\}$ be a sequence of non-negative real numbers with $\lim_{n\rightarrow\infty}a_n= a>0$. Prove that $$\lim_{n\rightarrow\infty}\sqrt{a_n} = \sqrt{a}$$

2) Let $\{X_n\}$ be a sequence of non-negative real numbers satisfying $\lim_{n\rightarrow\infty}X_n=0$ and $X_n\neq0$ for all $n\in N$. Prove that $$\lim_{n\rightarrow\infty}X_n \sin\left(\frac{1}{X_n}\right)=0$$

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    $\begingroup$ Welcome to Math.SE. What have you tried? We need a bit more information to push you in the right direction. $\endgroup$ – Nick Peterson Jun 23 '13 at 15:30
  • $\begingroup$ for #1 i try this: $\endgroup$ – tony Jun 23 '13 at 15:43
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    $\begingroup$ Let e > 0 let x=1/2 then a=<a^x + e)^1/x and a>(a^x-e)^1/x. since lim n-->infinite a_n = a there exist some N such that n>=N implies a_n<(a^x+e)^1/x. therefore n>=N we have a_n^x<a^x+e and a_n^x>a^x-e. it show that lim n-->infinite sqrt a_n=sqrt a $\endgroup$ – tony Jun 23 '13 at 15:49
  • $\begingroup$ @NicholasR.Peterson I think we can obtain a little more generous conclusion at this case. That is, a continuous function to the terms of a convergent sequence, the result is also convergent. $\endgroup$ – SundayCat Nov 10 '13 at 21:26
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Let me give you a couple of hints:

For #1: Note that $$ \sqrt{a_n}-\sqrt{a}=\frac{a_n-a}{\sqrt{a_n}+\sqrt{a}}, $$ by multiplying and dividing by $\sqrt{a_n}+\sqrt{a}$ and using the difference of squares formula.

For #2: Remember that $\sin(x)$ is bounded uniformly over all $x$. So, even though $\frac{1}{X_n}$ blows up, the sine term shouldn't make a big difference.

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Proposition: If $\lim_{n\rightarrow\infty}a_n = L$ and the function $f$ is continuous at $L$, then $$\lim_{n\rightarrow\infty}f(a_n) = f(L)$$

Proof: Since $\lim_{n\rightarrow\infty}a_n = L$,we know: $$\forall\eta \gt 0,~\exists N\gt0 ~~\text{such that }~~n\gt N, ~~\left|a_n - L\right|\lt\ \eta \tag{1}$$ Also,by the definition of continuity, we have: $$\forall\varepsilon \gt 0,~\exists \delta\gt0 ~~\text{such that }~~0\lt \left|x-L\right|\lt \delta, ~~\left|f(x) - f(L)\right|\lt\varepsilon\tag{2}$$ Let $\eta = \delta$, it guarantee that for any $\delta \gt 0$,we can always find an $N$, when $n\gt N,0\lt|a_n - L|<\delta$. Hence:

$$\forall\varepsilon\gt0, \exists N\gt0,\text{when}~~n\gt N,~0\lt|a_n - L|<\delta, ~~\text{such that},~~ |f(a_n) - f(L)| \lt \varepsilon\tag{3}$$ By $(3)$, we conclude $\lim_{n\rightarrow\infty}f(a_n) = f(L)$. Proved.


For you questions:

  1. Let $f(x) = \sqrt{x}$, it is easy to check that $\sqrt{x}$ is continuous at $a\ge 0$, hence apply the proposition above. we have $$\lim_{n\rightarrow\infty}f(a_n) = \lim_{n\rightarrow\infty}\sqrt{a_n}=f(L) = f(a) = \sqrt{a}$$

  2. Note that $-x \le x\sin\frac{1}{x} \le x$ and $\lim_{n\rightarrow\infty}-x_n = \lim_{n\rightarrow\infty}x_n = 0$.

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