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!

  • 1
    $\begingroup$ 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

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 $\sin x=-1$.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.