# Find this limit without L'hopital Rule : $\lim_{x\rightarrow +\infty}\frac{x(1+\sin(x))}{x-\sqrt{(1+x^2)}}$.

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!

-
No Limit.Try $x = k \pi,$ then try $x = \left( k + \frac{1}{2} \right) \pi.$ –  Will Jagy Oct 30 '13 at 0:16
@WillJagy: exactly! –  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 $(a-b)(a+b)=a^2-b^2$.