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Math Help - calculate limit

  1. #1
    Super Member dhiab's Avatar
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  2. #2
    Senior Member Spec's Avatar
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    There are a couple of different ways to show that the limit is 0.

    \lim_{x\to\infty}\frac{\ln(x^2+x+1)}{x}=\lim_{x\to  \infty}\frac{\ln\left(1+\frac{1}{x}+\frac{1}{x^2}\  right)+\ln x^2}{x}=\lim_{x\to\infty}\frac{\ln\left(1+\frac{1}  {x}+\frac{1}{x^2}\right)+2\ln x}{x}

    You should know that \lim_{x\to \infty}\frac{\ln x}{x}=0

    Or you can use L'Hospitals rule twice with f(x)=\ln(x^2+x+1) and g(x)=x

    \lim_{x\to \infty}\frac{f(x)}{g(x)}=\lim_{x\to \infty}\frac{f'(x)}{g'(x)}=\lim_{x\to \infty}\frac{f''(x)}{g''(x)}=\lim_{x\to \infty}\frac{2}{3x^2+2x+1}
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by spec View Post
    there are a couple of different ways to show that the limit is 0.

    \lim_{x\to\infty}\frac{\ln(x^2+x+1)}{x}=\lim_{x\to  \infty}\frac{\ln\left(1+\frac{1}{x}+\frac{1}{x^2}\  right)+\ln x^2}{x}=\lim_{x\to\infty}\frac{\ln\left(1+\frac{1}  {x}+\frac{1}{x^2}\right)+2\ln x}{x}

    you should know that \lim_{x\to \infty}\frac{\ln x}{x}=0

    or you can use l'hospitals rule twice with f(x)=\ln(x^2+x+1) and g(x)=x

    \lim_{x\to \infty}\frac{f(x)}{g(x)}=\lim_{x\to \infty}\frac{f'(x)}{g'(x)}=\lim_{x\to \infty}\frac{f''(x)}{g''(x)}=\lim_{x\to \infty}\frac{2}{3x^2+2x+1}
    thank you
    if x to -infty ?
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  4. #4
    Senior Member Spec's Avatar
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    Same result.
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