[SOLVED] Help finding limits that approach ∞.

• Oct 27th 2008, 03:49 PM
ilikedmath
[SOLVED] Help finding limits that approach ∞.

I rationalized the expression by multiplying the numerator and denominator by the conjugate of the denominator, but then I got stuck and didn't know where to go from there. It seems that my rationalizing just made finding the limit more complicated.(Thinking) Would I then divide everything by what dominates the denominator? So "n" dominates the denominator so if I divide everything by n, I get: http://www.cramster.com/Answer-Board...1875008028.gif. I know that (1/n) goes to 0, but how can I get rid of the '3n'?

I can rewrite (4^n)/(7^n) as (4/7)^n. And since (4/7)^n goes to 0, the expression further simplifies to:
http://www.cramster.com/Answer-Board...8125001062.gif. But then if (4/7)^n goes to 0, I get 11/0 which is undefined. So I'm wrong. (Headbang)

Do I need to use the squeeze theorem for these? I haven't had calculus in two years, so I am very rusty. This is for a intro to real analysis class.

Any help, tips, corrections, and/or suggestions are greatly appreciated.

- Stumped Student(Nerd)
• Oct 27th 2008, 04:56 PM
Soroban
Hello, ilikedmath!

Quote:

$1)\;\;\lim_{n\to\infty}\frac{\sqrt{3n-1}}{\sqrt{n}+2}$
Divide top and bottom by $\sqrt{n}$

$\lim_{n\to\infty}\left(\frac{\dfrac{\sqrt{3n-1}}{\sqrt{n}}} {\dfrac{\sqrt{n}}{\sqrt{n}} + \dfrac{2}{\sqrt{n}}}\right) \;=\;\lim_{n\to\infty}\left( \frac{\sqrt{\dfrac{3n}{n} - \dfrac{1}{n}}} {1 + \dfrac{2}{\sqrt{n}}}\right)$ . $=\;\lim_{n\to\infty}\left(\frac{\sqrt{3 - \dfrac{1}{n}}}{1 + \dfrac{2}{\sqrt{n}}}\right) \;=\;\frac{\sqrt{3-0}}{1+0} \;=\;\sqrt{3}$

Quote:

$2)\;\;\lim_{n\to\infty}\frac{4^{n+1} + 7^{n+1}}{4^n + 7^n}$
Divide top and bottom by $7^n$

$\lim_{n\to\infty}\left[ \frac{\dfrac{4^{n+1}}{7^n} + \dfrac{7^{n+1}}{7^n}} {\dfrac{4^n}{7^n} + \dfrac{7^n}{7^n}}\right] \;=\;\lim_{n\to\infty}\left[ \frac{4\left(\dfrac{4}{7}\right)^n + 7} {\left(\dfrac{4}{7}\right)^n + 1}\right]$ . $= \;\frac{4\!\cdot\!0 + 7}{0 + 1} \;=\;7$

• Oct 27th 2008, 07:01 PM
ilikedmath
Quote:

Originally Posted by Soroban
Hello, ilikedmath!

(Giggle) <-- me laughing at myself for making it a lot more complicated than it was.

Thank you very much!!! Very clear and helpful answer you gave, A+++! (Cool)