solve sqrt(9n^2+6n+2)-3n whit lim.; 2^n+1-3_n+5_n-1/2^n+3^n+

[spepe]

I am new to this forum (just registered) But I need some help! :S I just don't understand how to solve these whit lim.

sqrt(9n[sup:204j5ms3]2[/sup:204j5ms3]+6n+2)-3n

2[sup:204j5ms3]n+1[/sup:204j5ms3]-3[sub:204j5ms3]n[/sub:204j5ms3]+5[sub:204j5ms3]n-1[/sub:204j5ms3]/2[sup:204j5ms3]n[/sup:204j5ms3]+3[sup:204j5ms3]n+1[/sup:204j5ms3]+5[sup:204j5ms3]n-2[/sup:204j5ms3]

Sorry for my English!

Bye!

 

[stapel]

I just don't understand how to solve these whit lim.

I'm sorry, but I don't know what a "whit lim" might be...? Also, you refer to "solving", but you have not posted equations.

Please reply with clarification. Thank you!

Eliz.

 

[soroban]

Hello, spepe!

Welcome aboard!

I'll take a guess at what you meant.

I assume the limit is: \(\displaystyle n \to \infty}\)

$\displaystyle \lim_{n\to\infty}\bigg(\sqrt{9n^2 - 6n + 2} - 3n\bigg)$

$\displaystyle \text{Multiply top and bottom by }\left(\sqrt{9n^2-6n+2} + 3n\right)$

. . $\displaystyle \frac{\sqrt{9n^2-6n+2} - 3n}{1}\cdot\frac{\sqrt{9n^2-6n+2} + 3n}{\sqrt{9n^2-6n+2} + 3n} \;=\;\frac{(9n^2-6n+2) - 9n^2}{\sqrt{9n^2-6n+2} + 3n} \;=\;\frac{-6n + 2}{\sqrt{9n^2-6n+2} + 3n}$


Divide top and bottom by $\displaystyle n$:

. . $\displaystyle \frac{\frac{-6n}{n} + \frac{2}{n}} {\frac{\sqrt{9n^2-6n+2}}{n} + \frac{3n}{n}} \;=\; \frac{-6 + \frac{2}{n}}{\sqrt{\frac{9n^2}{n^2} - \frac{6n}{n^2} + \frac{2}{n^2}}+ 3} \;=\;\frac{-6+\frac{2}{n}} {\sqrt{9 - \frac{6}{n} + \frac{2}{n^2}} + 3}$


$\displaystyle \text{Therefore: }\;\lim_{n\to\infty}\left[\frac{-6+\frac{2}{n}} {\sqrt{9 - \frac{6}{n} + \frac{2}{n^2}} + 3}\right] \;=\;\frac{-6 + 0}{\sqrt{9 - 0 + 0} + 3} \;=\;\frac{\text{-}6}{6} \;=\;-1$

$\displaystyle \lim_{n\to\infty} \frac{2^{n+1} - 3^n + 5^n}{2^n + 3^{n+1} + 5^{n-2}}$

$\displaystyle \text{Fact . . .}\;\;\text{For }r < 1\!:\;\;\lim_{n\to\infty} r^n \:=\:0$


$\displaystyle \text{Divide top and bottom by }5^{n+1}$

. . $\displaystyle \lim_{n\to\infty}\left[\frac{\frac{2^{n+1}}{5^{n+1}} - \frac{3^n}{5^{n+1}} + \frac{5^n}{5^{n+1}} } {\frac{2^n}{5^{n+1}} + \frac{3^{n+2}}{5^{n+1}} + \frac{5^{n-2}}{5^{n+1}}}\right] \;=\;\lim_{n\to\infty}\left[\frac{\left(\frac{2}{5}\right)^{n+1} - \frac{1}{5}\left(\frac{3}{5}\right)^n + \frac{1}{5}} {\frac{1}{5}\left(\frac{2}{5}\right)^n + \left(\frac{3}{5}\right)^{n+1} + \frac{1}{5^3}}\right]$

. . $\displaystyle =\; \frac{0 - \frac{1}{5}\!\cdot\!0 + \frac{1}{5}}{\frac{1}{5}\!\cdot\!0 + 0 + \frac{1}{5^3}} \;=\;\frac{\frac{1}{5}}{\frac{1}{125}} \;=\;25$

 

[mmm4444bot]

2[sup:3ajgngis]n+1[/sup:3ajgngis]-3[sub:3ajgngis]n[/sub:3ajgngis]+5[sub:3ajgngis]n-1[/sub:3ajgngis]/2[sup:3ajgngis]n[/sup:3ajgngis]+3[sup:3ajgngis]n+1[/sup:3ajgngis]+5[sub:3ajgngis]n-2[/sub:3ajgngis]

$\displaystyle \frac{2^{n+1} - 3^n + 5^n}{2^n + 3^{n+1} + 5^{n-2}}$

Soroban interprets the wave, and almost comes out dry. Way to go :!: