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Thread: A couple of limits to solve

  1. #1
    Dec 2007

    Question A couple of limits to solve

    Stumped on two problems:
    1. \lim_{n\to\infty}\frac{8^n(2*4^n+3*5^n)}{5^n(2*7^n-3*8^n)}<br />
    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. \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! =)
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  2. #2
    MHF Contributor red_dog's Avatar
    Jun 2007
    Medgidia, Romania
    1) \lim_{n\to\infty}\frac{8^n(2\cdot 4^n+3\cdot 5^n)}{5^n(7^n-3\cdot 8^n)}=

    =\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]}=

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

    =\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)}=

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

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