Results 1 to 2 of 2

Thread: A couple of limits to solve

  1. #1
    Dec 2007

    Question 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! =)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Jun 2007
    Medgidia, Romania
    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) $\displaystyle \lim_{x\to 4}\frac{\sqrt{1+2x}-3}{\sqrt{x}-2}=$

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

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

    $\displaystyle =\lim_{x\to 4}\frac{2(\sqrt{x}+2)}{\sqrt{1+2x}+3}=\frac{8}{6}= \frac{4}{3}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How to solve limits?
    Posted in the Calculus Forum
    Replies: 23
    Last Post: Dec 14th 2010, 09:04 PM
  2. Limits problem , i cant solve
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: Nov 7th 2009, 10:02 PM
  3. A Couple Of Limits
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Dec 5th 2008, 09:42 AM
  4. Replies: 2
    Last Post: Nov 30th 2008, 08:01 PM
  5. Replies: 5
    Last Post: Nov 20th 2006, 12:54 PM

Search Tags

/mathhelpforum @mathhelpforum