Proof of Limits Code Answer

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Prove that:

$$limlimits_{n to infty} frac{1}{sqrt[4]{n}} = 0$$

Using the formal definition of convergence.


$frac{1}{sqrt[5]{n}} > frac{1}{sqrt[4]{n}}$

To make $frac{1}{sqrt[5]{n}}$ less than $epsilon$.

Need $frac{1}{sqrt[5]{n}}$ < $epsilon$

That is $frac{1}{sqrt[5]{epsilon}} < n$

So let $N = frac{1}{sqrt[5]{n}}$


Suppose $epsilon > 0$

Let $N = frac{1}{sqrt[5]{n}}$

For any $n > N$ we have that $n > frac{1}{sqrt[5]{epsilon}}$.

And we have:

$|s_n – L| = left|frac{1}{sqrt[4]{n}} – 0right|$

$|s_n – L| = left|frac{1}{sqrt[4]{n}}right|$

$|s_n – L| = frac{1}{sqrt[4]{n}}$

$leq frac{1}{sqrt[5]{n}}$

$< frac{1}{N}$

$leq frac{1}{frac{1}{epsilon}}$

$= epsilon$


$$limlimits_{n to infty} frac{1}{sqrt[4]{n}} = 0$$

I’m not sure if I did this correctly. Any help will be appreciated.


It doesn’t make sense to take $N = 1/sqrt[5]{n}$, because it depends on the index of the sequence and not on $epsilon$. Instead, let $N$ be an integer greater than $1/epsilon^4$. Then for $n ge N$,

$$frac{1}{sqrt[4]{n}} le frac{1}{sqrt[4]{N}} < frac{1}{sqrt[4]{1/epsilon^4}} = frac{1}{1/epsilon} = epsilon.$$

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