# Thread: proving a limit by definition..

1. ## proving a limit by definition..

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

2. but when X is big enought, 2X is big too, but sin(3X) remains between -1 and 1... doesn´t it lead you to somewhere?

3. yes it leads to infinity but thats only word not a proof by definition like i showed

4. Surely you can show that for $\displaystyle x\ge 2,$

$\displaystyle \sqrt {2x - 3} < \sqrt {2x - \sin (3x)} < \sqrt {2x + 3}~?$

5. Plato just showed what I meant ^^