# A couple of limits to solve

• Sep 17th 2009, 10:19 PM
xxlvh
A couple of limits to solve
Stumped on two problems:
1. $\displaystyle \lim_{n\to\infty}\frac{8^n(2*4^n+3*5^n)}{5^n(2*7^n-3*8^n)}$
I haven't worked with limits in awhile, and I can't find any examples in my notes involving exponentials.

The second problem is:
2. $\displaystyle \lim_{x\to 4}\frac{\sqrt{1+2x}-3}{\sqrt{x}-2}$
Rationalizing didnt work out very well, my preferred method would have been L Hopitals rule but we havent covered it in class yet so Im not sure if I can use that in my work.

Any help would be greatly appreciated! =)
• Sep 18th 2009, 01:58 AM
red_dog
1) $\displaystyle \lim_{n\to\infty}\frac{8^n(2\cdot 4^n+3\cdot 5^n)}{5^n(7^n-3\cdot 8^n)}=$

$\displaystyle =\lim_{n\to\infty}\frac{8^n\cdot 5^5\left[2\left(\frac{4}{5}\right)^n+3\right]}{5^n\cdot 8^n\left[\left(\frac{7}{8}\right)^n-3\right]}=$

$\displaystyle =\lim_{n\to\infty}\frac{2\left(\frac{4}{5}\right)^ n+3}{\left(\frac{7}{8}\right)^n-3}=\frac{3}{-3}=-1$

2) $\displaystyle \lim_{x\to 4}\frac{\sqrt{1+2x}-3}{\sqrt{x}-2}=$

$\displaystyle =\lim_{x\to 4}\frac{(\sqrt{1+2x}-3)(\sqrt{1+2x}+3)}{\sqrt{1+2x}-3}\cdot\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}=$

$\displaystyle =\lim_{x\to 4}\frac{2(x-4)}{\sqrt{1+2x}+3}\cdot\frac{\sqrt{x}+2}{x-4}=$

$\displaystyle =\lim_{x\to 4}\frac{2(\sqrt{x}+2)}{\sqrt{1+2x}+3}=\frac{8}{6}= \frac{4}{3}$