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How can I evaluate the limit

$$\lim_{x \to \infty} \frac{1}{x(x+1)}$$

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    $\begingroup$ $$0\le\lim_{x\to\infty}{1\over x(x+1)}\le\lim_{x\to\infty}{1\over x}=0$$ $\endgroup$ Jul 22, 2014 at 22:17
  • $\begingroup$ What are your thoughts? $\endgroup$
    – hardmath
    Jul 22, 2014 at 22:19
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    $\begingroup$ When $x$ is big, what is $\frac{1}{x(x+1)}$ close to? You undoubtedly know. $\endgroup$ Jul 22, 2014 at 22:27
  • $\begingroup$ I can see how it would be close to 0, just curious as to how to evaluate it completely $\endgroup$
    – user8028
    Jul 22, 2014 at 22:35
  • $\begingroup$ What's wrong with the question? $\endgroup$ Jul 23, 2014 at 0:20

1 Answer 1

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Although it's pretty obvious that $x(x+1)=x^2+x$ goes to infinity for $x\to\infty$, you can use partial fraction decomposition if it is not clear: $$\lim_{x\to\infty}{1\over x(x+1)}=\lim_{x\to\infty}\left({1\over x}-{1\over x+1}\right)=0$$

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  • $\begingroup$ So I just set up $\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{1+x} $ $\endgroup$
    – user8028
    Jul 22, 2014 at 22:40
  • $\begingroup$ And in this case A and B will both = 1? $\endgroup$
    – user8028
    Jul 22, 2014 at 22:40
  • $\begingroup$ $A=1$ and $B=-1$. You can obtain it by setting $$A(x+1)+Bx=1$$ After that compare coefficients and solve the system$$A+B=0\\A=1$$ $\endgroup$ Jul 22, 2014 at 22:47

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