# Evaluate the limit $\lim \limits_{x \to \infty} \frac{1}{x(x+1)}$ [closed]

How can I evaluate the limit

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

• $$0\le\lim_{x\to\infty}{1\over x(x+1)}\le\lim_{x\to\infty}{1\over x}=0$$ – Adam Hughes Jul 22 '14 at 22:17
• What are your thoughts? – hardmath Jul 22 '14 at 22:19
• When $x$ is big, what is $\frac{1}{x(x+1)}$ close to? You undoubtedly know. – André Nicolas Jul 22 '14 at 22:27
• I can see how it would be close to 0, just curious as to how to evaluate it completely – user8028 Jul 22 '14 at 22:35
• What's wrong with the question? – Vishwa Iyer Jul 23 '14 at 0:20

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: $$\lim_{x\to\infty}{1\over x(x+1)}=\lim_{x\to\infty}\left({1\over x}-{1\over x+1}\right)=0$$

• So I just set up $\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{1+x}$ – user8028 Jul 22 '14 at 22:40
• And in this case A and B will both = 1? – user8028 Jul 22 '14 at 22:40
• $A=1$ and $B=-1$. You can obtain it by setting $$A(x+1)+Bx=1$$ After that compare coefficients and solve the system$$A+B=0\\A=1$$ – Dmitry Kazakov Jul 22 '14 at 22:47