Help verifying $\lim \frac{2n^2}{n^3 +3} = 0$

Question

I am having difficulty with verifying the following using the definition of convergence of a sequence(i.e. a sequence converges to $x$, if $n \ge N$ $\Longrightarrow$ $|x-x_n|< \epsilon$ )

$$\lim \frac{2n^2}{n^3 +3} = 0$$

Now if I take $$\left|\frac{2n^2}{n^3 +3} - 0\right|< \epsilon$$

I get stuck at $$\frac{n^2}{n^3 +3}< \frac{\epsilon}{2}$$ If I am not wrong we have to show $n>A \epsilon$ for some expression $A$ and merely writing $$n<\frac{\sqrt \epsilon \sqrt {n^3 +3}} {2}$$ would be incorrect right because we have to show\write $\epsilon$ independently of $n$?

I am having the same difficulty simplifying the relationship between $\epsilon$ and $n$ in the following expression

$$\left|\frac{\sin (n^2)}{\sqrt[3] n}- 0\right| < \epsilon$$

Can anyone give me a hint for these?

Answer 1

Note that $ \frac{2n^2}{n^3 +3} \lt \frac{2n^2}{n^3} =\frac{2}{n} $.

Therefore, if $\frac{2}{n} \lt \epsilon $ then $ \frac{2n^2}{n^3 +3} \lt \epsilon $.

You don't have to get the best $n$ - a fairliy good one is enough to prove convergence.

Answer 2

Hints $$ 0\le\left\lvert\frac{\sin (n^2)}{n^{1/3}}\right\rvert\leq \frac{1}{n^{1/3}} $$ and $$ 0\le\frac{2n^2}{n^3+3}\leq 2\frac{n^2}{n^3}=\frac{2}{n}. $$

Answer 3

Hint:

For $n > \frac2{\varepsilon}$ we have

$$\frac{n^2}{n^3+3} < \frac{n^2}{n^3} = \frac1n < \frac\varepsilon2$$