1. find the limit 2. use the definition of the limit to prove that

**kawafis44** · November 3rd 2007, 04:21 AM

1. Find the limit
$\lim{n\to\infty}\sqrt{n+2\sqrt{n}}-\sqrt{n}$

2. Use the definition of the limit to prove that
$\lim_{n\to\infty}\frac{3n^2-1}{n+2}=+\infty$
(I try to solve it like this $|\frac{3n^2-1}{n+2}-\infty|=\varepsilon$ but what then?)

**kalagota** · November 3rd 2007, 06:27 AM

Originally Posted by kawafis44

1. Find the limit
$\lim{n\to\infty}\sqrt{n+2\sqrt{n}}-\sqrt{n}$

2. Use the definition of the limit to prove that
$\lim_{n\to\infty}\frac{3n^2-1}{n+2}=+\infty$
(I try to solve it like this $|\frac{3n^2-1}{n+2}-\infty|=\varepsilon$ but what then?)

1)
$\lim_{n\rightarrow\infty} \sqrt{n+2\sqrt{n}}-\sqrt{n} \cdot \frac{\sqrt{n+2\sqrt{n}}+\sqrt{n}}{\sqrt{n+2\sqrt{ n}}+\sqrt{n}}$

then use L'hospital's Rule..

2)
$\lim_{n\rightarrow\infty}\frac{3n^2-1}{n+2}=+\infty$

this is the definition for that case right?

$\lim_{x\rightarrow\infty} f(x) = \infty \Longleftrightarrow \forall \alpha \in R, \exists \beta \in R$ such that if $x > \beta \implies f(x) > \alpha.$

also, $\frac{3n^2-1}{n+2} = 3n - 6 + \frac{11}{n+2}$
so that,
$\lim_{n\rightarrow\infty}\frac{3n^2-1}{n+2} = \lim_{n\rightarrow\infty} 3n - \lim_{n\rightarrow\infty}6 + \lim_{n\rightarrow\infty}\frac{11}{n+2}$

from here, you only need to show that $\lim_{n\rightarrow\infty} 3n = \infty$

so, if $\alpha \in R$ , let $\beta := sup \left\{1, \alpha \right\}$ . Then for all $n > \beta$ , we have $3n > n > \alpha$ .

**Krizalid** · November 3rd 2007, 06:54 AM

Originally Posted by kawafis44

1. Find the limit
$\lim_{n\to\infty}\sqrt{n+2\sqrt{n}}-\sqrt{n}$

For this one, we don't need L'Hôpital's Rule. After multiplyin' the expression, the numerator will be $2\sqrt n,$ and we can handle the limit easily.

**kalagota** · November 3rd 2007, 06:57 AM

Originally Posted by Krizalid

For this one, we don't need L'Hôpital's Rule. After multiplyin' the expression, the numerator will be $2\sqrt n,$ and we can handle the limit easily.

actually, i really don't see it easily.. Ü can you show it.. thanks.. Ü

**Krizalid** · November 3rd 2007, 07:02 AM

Originally Posted by kalagota

1)
$\lim_{n\rightarrow\infty} \sqrt{n+2\sqrt{n}}-\sqrt{n} \cdot \frac{\sqrt{n+2\sqrt{n}}+\sqrt{n}}{\sqrt{n+2\sqrt{ n}}+\sqrt{n}}$

I'll take it from here:

$\frac{{\left( {\sqrt {n + 2\sqrt n } - \sqrt n } \right)\left( {\sqrt {n + 2\sqrt n } + \sqrt n } \right)}} {{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{{n + 2\sqrt n - n}} {{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{{2\sqrt n }} {{\sqrt {n + 2\sqrt n } + \sqrt n }}.$

Now, divide by $\sqrt n$ top & bottom, so

$\frac{{2\sqrt n }} {{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{2} {{\sqrt {1 + \dfrac{2} {{\sqrt n }}} + 1}}.$

The conclusion follows.

--

By the way, applyin' L'Hôpital's Rule, it's a mess, with all those roots.

**kalagota** · November 3rd 2007, 07:05 AM

Originally Posted by Krizalid

I'll take it from here:

$\frac{{\left( {\sqrt {n + 2\sqrt n } - \sqrt n } \right)\left( {\sqrt {n + 2\sqrt n } + \sqrt n } \right)}} {{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{{n + 2\sqrt n - n}} {{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{{2\sqrt n }} {{\sqrt {n + 2\sqrt n } + \sqrt n }}.$

Now, divide by $\sqrt n$ top & bottom, so

$\frac{{2\sqrt n }} {{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{2} {{\sqrt {1 + \dfrac{2} {{\sqrt n }}} + 1}}.$

The conclusion follows.

--

By the way, applyin' L'Hôpital's Rule, it's a mess, with all those roots.

ahh limit as the variable approaches infinity, i forgot it.. hahaha, maybe i'm just tired.. thanks!!

**kawafis44** · November 3rd 2007, 08:11 AM

ok, thanks very much

how to find this lim ?
$\lim_{n\to\infty}\sqrt[3]{n^3+9}-n$

**Krizalid** · November 3rd 2007, 08:25 AM

Originally Posted by kawafis44

how to find this lim ?
$\lim_{n\to\infty}\sqrt[3]{n^3+9}-n$

The trick is, to apply the identity $a^3-b^3=(a-b)(a^2+ab+b^2).$

Let's just focus on $\sqrt[3]{n^3+9}-n.$ Call $u=\sqrt[3]{n^3+9}.$

From the expression $u-n,$ we'll multiply and divide by $u^2+nu+n^2,$ so $u-n=\frac{u^3-n^3}{u^2+nu+n^2}.$

Now back substitute and yields

$\sqrt[3]{n^3+9}\,-\,n=\frac{n^3+9-n^3}{\sqrt[3]{(n^3+9)^2}+n\sqrt[3]{n^3+9}+n^2}=\frac9{\sqrt[3]{(n^3+9)^2}+n\sqrt[3]{n^3+9}+n^2}.$

From here, clearly, the limit is 0.

**kawafis44** · November 3rd 2007, 12:18 PM

Thanks, Krizalid. Now I see that Chile has similar flag to Poland

.
Now I've got problem with something like that:
24. $a_{n} = \sqrt[2n]{1+2^{-n}+3(0.2)^{n}}$

**Plato** · November 3rd 2007, 12:46 PM

If you know the following the limit is obvious.
$n \ge 4 \Rightarrow \quad 1 \le \sqrt[{2n}]{{1 + 2^{ - n} + 3\left( {.2} \right)^n }} \le \sqrt[{2n}]{3} $

**kawafis44** · November 3rd 2007, 01:13 PM

Thanks, I try to solve it (to find lim):
18. $a_{n} = \frac{nsin\frac{1}{n}}{n\sqrt{n}+2} = \frac{1cos\frac{1}{n}}{1\sqrt{n}+n\frac{1}{2\sqrt{ n}}}$
but $lim_{n}n\frac{1}{2\sqrt{n}} = \infty*0$ is undefined

**Plato** · November 3rd 2007, 02:13 PM

Because $n\sin \left( {\frac{1}{n}} \right) = \frac{{\sin \left( {\frac{1}{n}} \right)}}{{\frac{1}{n}}} \to 0$ you have a bounded numerator.
So what can be said of the limit?

**kawafis44** · November 4th 2007, 12:40 AM

so how to solve it?

**Jhevon** · November 4th 2007, 12:59 AM

Originally Posted by kawafis44

so how to solve it?

are you talking about $a_n = \frac {n \sin \left( \frac 1n \right)}{n \sqrt{n} + 2}$ ?

if so, divide through the numerator and denominator by $n$

you get that $a_n = \frac {\sin \left( \frac 1n \right)}{\sqrt{n} + \frac 2n}$

now follow Plato's hint. the numerator is bounded, and the denominator will increase without bound

**kawafis44** · November 4th 2007, 03:09 AM

thanks, now limes of sequences with trigonometry functions are more clear for me

. but what about complex numbers? i have got example like that to solve:
30. $a_{n} = \frac{in^2+2}{n^2-n+1}$

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