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Find this limit without l'Hopital rule : $$\lim_{x\rightarrow +\infty}\frac{x(1+ \sin x)}{x-\sqrt{1+x^2}}$$

I tried much but can't get any progress!

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    $\begingroup$ No Limit.Try $x = k \pi,$ then try $x = \left( k + \frac{1}{2} \right) \pi.$ $\endgroup$ – Will Jagy Oct 30 '13 at 0:16
  • $\begingroup$ @WillJagy: exactly! $\endgroup$ – Iloveyou Dec 4 '13 at 14:39
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The limit does not exist. Multiply top and bottom by $x+\sqrt{1+x^2}$. The bottom becomes $-1$. As to the new top, it is very big if $\sin x$ is not close to $-1$. However, there are arbitrarily large $x$ such that $\sin x=-1$.

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Amplify both sides with $x+\sqrt{1+x^2}$ , and use the fact that $(a-b)(a+b)=a^2-b^2$.

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