$$\lim_{x\to\infty} x\left(\sqrt{x^2+2x}-2\sqrt{x^2+x}+x\right) =\ ? $$
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The sequence $a_n=n$ is divergent. So $\lim a_nb_n$ will exist only if $b_n\to 0$. So, you have to check if the bracketed term converges to zero. If no, then the limit won't exist. – Jul 23 '16 at 11:51
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The limit here exists and is equal to $-1/4$. – smcc Jul 23 '16 at 11:55
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You might have a look at answers to this question to see whether you can use a similar approach. – Martin Sleziak Jul 23 '16 at 12:09
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Take $t=1/x$, then the limit becomes $$\lim_{t\to 0^+} \frac{1}{t^2}\left(\sqrt{1+2t}-2\sqrt{1+t}+1\right).$$ Then use $\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+o(x^2)$. $$\lim_{t\to 0^+} \frac{1}{t^2}\left(\left(1+\frac{(2t)}{2}-\frac{(2t)^2}{8}+o(t^2)\right)-2\left(1+\frac{t}{2}-\frac{t^2}{8}+o(t^2)\right)+1\right)=-\frac{1}{4}.$$
Robert Z
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