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

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

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?)

2. 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 $3n > n > \alpha$.

3. 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.

4. 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

5. 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.

6. 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!!

7. ok, thanks very much
how to find this lim ?
$\lim_{n\to\infty}\sqrt[3]{n^3+9}-n$

8. 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.

9. Thanks, Krizalid. Now I see that Chile has similar flag to Poland .
Now I've got problem with something like that:
24. $a_{n} = \sqrt[2n]{1+2^{-n}+3(0.2)^{n}}$

10. 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}
$

11. 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

12. 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?

13. so how to solve it?

14. 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

15. 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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