$\displaystyle \lim_{x\rightarrow\infty}\sqrt{2x-\sin3x}=\infty$

we need to prove that for every N>0 there is M>0 so for every x>M

we have f(x)>N .

i have a solution of a similar question:

$\displaystyle \lim_{x\rightarrow\infty}x^{2}+3=\infty$

the solution is:

we take $\displaystyle N\geq3$ we find M that fits it.

we clame that M=$\displaystyle \sqrt{N-3}$

if x>M then x>$\displaystyle \sqrt{N-3}$

so $\displaystyle x^{2}+3>N$

i cant use the similar question because i cantisolate x on one side.

and i cant use anything but thisdefinition to solve it