Krdytkyu

I am interested in the limit

$$ lim_{n rightarrow infty} sqrt[n]{n^3}$$

Can we simply conclude that:
$$ lim_{n rightarrow infty} (sqrt[n]{n})^3= 1^3=1.$$
I have proven that $sqrt[n]{n}rightarrow1$ earlier in this textbook. Also since the limit of a power is the power of the limit.

Yes, we can simply do that. Since the exponent $^3$ is a constant neutral number (meaning we may interpret it as a fixed number of multiplications) we can move the limit inside of it. So if you already know $lim_{ntoinfty}sqrt[n]n=1$ then that's a full proof.

Yes we are allowed to do that since for continuity

$$lim_{xto x_0} f(x)=Lin mathbb{R} implies lim_{xto x_0} [f(x)]^k=left[lim_{xto x_0} [f(x)right]^k=L^k$$

Indeed in that particular case since $sqrt[n]{n}to 1$

$$forall epsilon>0 quad exists bar n quad forall n>bar n quad |sqrt[n]{n}-1|<epsilon$$

we have that, assuming $sqrt[n]{n}<2$, $forall bar epsilon=7epsilon>0$

$$|(sqrt[n]{n})^3-1|=|sqrt[n]{n}-1|cdot |sqrt[n]{n^2}+sqrt[n]{n}+1|< 7epsilon=bar epsilon quad forall n>bar n$$

and then $(sqrt[n]{n})^3to 1$.

You know that
$sqrt[n]{n}
to 1$.

Also
$sqrt[n]{n^k}
=(sqrt[n]{n})^k
$
so

$begin{array}\
sqrt[n]{n^k}-1
&=(sqrt[n]{n})^k-1\
&=(sqrt[n]{n}-1)sum_{j=0}^{k-1} sqrt[n]{n}^j\
text{so}\
|sqrt[n]{n^k}-1|
&=|(sqrt[n]{n}-1)sum_{j=0}^{k-1} sqrt[n]{n}^j|\
&=|(sqrt[n]{n}-1)||sum_{j=0}^{k-1} sqrt[n]{n}^j|\
&le|(sqrt[n]{n}-1)|sum_{j=0}^{k-1} |sqrt[n]{n}^j|\
&le|(sqrt[n]{n}-1)|sum_{j=0}^{k-1} |sqrt[n]{n}^n|\
&=|(sqrt[n]{n}-1)|sum_{j=0}^{k-1} |n|\
&=|(sqrt[n]{n}-1)|kn\
end{array}
$

Therefore,
to make
$|sqrt[n]{n^k}-1|
le epsilon$,
choose $n$ large enough
so that
$|(sqrt[n]{n}-1)|
le frac{epsilon}{kn}
$.

You can if you know four things.

1) If $f$ is continuous and $a_n to M$ and for all $a_n$ that $f(a_n)$ is defined and $f(M)$ is defined then $a_nto M$ means $f(a_n) to f(M)$.

But you DO have to prove this sometime. I'm sure your text has proven that somewhere.

And

2) you need to know that $sqrt[n]{n} to 1$.

But you claim you have already shown that.

And 3) that $sqrt[n]{n^3} = (sqrt[n]{n})^3$.

Which is basic. It follows that as for all $M > 0$ and $nin mathbb N$ there is a unique $k = sqrt[n]{M}$ so that $k^n = M$. And as $(k^j)^n= (k^n)^j = M^j; for j in mathbb N$ it follows that $(sqrt[n]{M})^j= sqrt[n]{M^3}$.

And finally you need to know 4) that $()^3: mathbb R to mathbb R$ is continuous.

Which is .... basic. But you should have proven that sometime.

So $lim sqrt[n]{n^3} = lim (sqrt[n]{n})^3 = (lim sqrt[n]{n})^3 = 1^3 =1$.

Yes, because the cubic root function is continuous, so that you can swap the limit and the root.

Continuity at $1$ is ensured by the fact that

$$|sqrt[3]{1+delta}-1|=left|frac{delta}{(sqrt[3]{1+delta})^2+sqrt[3]{1+delta}+1}right|<fracdelta3<epsilon$$ holds with $delta<3epsilon$.