proving a limit by definition..

**transgalactic** · November 9th 2010, 06:43 AM

$\lim_{x\rightarrow\infty}\sqrt{2x-\sin3x}=\infty$
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:
$\lim_{x\rightarrow\infty}x^{2}+3=\infty$
the solution is:
we take $N\geq3$ we find M that fits it.
we clame that M= $\sqrt{N-3}$
if x>M then x> $\sqrt{N-3}$
so $x^{2}+3>N$
i cant use the similar question because i cantisolate x on one side.

and i cant use anything but thisdefinition to solve it

**teachermath** · November 9th 2010, 06:49 AM

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?

**transgalactic** · November 9th 2010, 06:52 AM

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

**Plato** · November 9th 2010, 06:54 AM

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

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

**teachermath** · November 9th 2010, 08:12 AM

Plato just showed what I meant ^^

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