Prove that the sequence $\frac{\sin(k)+k^2}{\root\of{1+3k^4}}$ converges using the $\epsilon$ criteria.

Question

Propose that the limit $L = \frac{1}{\root\of3}$.

Given $\epsilon \gt 0 \exists k_\epsilon$ such that for all $k \gt k_\epsilon$ we have $\lvert\frac{\sin(k)+k^2}{\root\of{1+3k^4}} - \frac{1}{\root\of3}\rvert \lt \epsilon$. We have that $\lvert\frac{\sin(k)+k^2}{\root\of{1+3k^4}} - \frac{1}{\root\of3}\rvert\leq \lvert\frac{1+k^2}{\root\of{1+3k^4}} - \frac{1}{\root\of3}\rvert$ as $\sin$ is bounded above by $1$. $\lvert\frac{1+k^2}{\root\of{1+3k^4}} - \frac{1}{\root\of3}\rvert \lt 1+k^2 = \epsilon$.

I'm pretty sure the last inequality is wrong but I'm not certain could somebody state whether it is wrong and how I should proceed if I am wrong.

Answer 1

First, noting that $|\sin(k)|\le 1$, then given $\epsilon>0$ we have

$$\left|\frac{\sin(k)}{\sqrt{1+3k^4}}\right|\le \frac{1}{k^2}<\frac{\epsilon}{2}$$

whenever, $k>K_1=\sqrt{\frac{2}{\epsilon}}$.

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Next, we have

$$\left|\frac{k^2}{\sqrt{1+3k^4}}-\frac1{\sqrt3}\right|=\left|\frac{1}{\left(\sqrt3k^2+\sqrt{1+3k^4}\right)\sqrt{3}\sqrt{1+3k^4}}\right|\le \frac{1}{k^4}<\frac{\epsilon}{2}$$

whenever $k>K_2=\sqrt[4]{\frac{2}{\epsilon}}$.

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Finally, using the triangle inequality, we see that given $\epsilon>0$,

$$\left|\frac{\sin(k)+k^2}{\sqrt{1+3k^4}}-\frac1{\sqrt3}\right|\le \left|\frac{\sin(k)}{\sqrt{1+3k^4}}\right|+\left|\frac{k^2}{\sqrt{1+3k^4}}-\frac1{\sqrt3}\right|<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon$$

whenever $k>\max(K_1,K_2)$

Answer 2

Also, it may be easier to show that $\lim_{k \to \infty}\left(\frac{k^2}{\sqrt{1+3k^4}}\right)^2 = \lim_{k \to \infty}\frac{k^4}{1+3k^4} = \frac{1}{3}$, from which it follows that $\lim_{k \to \infty}\frac{k^2}{\sqrt{1+3k^2}} = \frac{1}{\sqrt{3}}$. This would plug into Dr. MV's answer, but save you a bit of ugly algebra.