1
$\begingroup$

I want to evaluate the given limit using the Sandwich theorem. $$\lim_{n\to\infty}k^{\frac{1}{n}} \hspace{20pt} ,\forall\,k>0$$ I know that this limit evaluates to 1 but I can't find the bounding function for it such that: $$f(x)\le k^\frac{1}{n} \le g(x)$$ Please give me an insight on how to approach such situations.

$\endgroup$
2
  • $\begingroup$ your upper limit is wrong. For any $k > 1$, $1^{1/n}$ will be less than $k^{1/n}$ $\endgroup$ – Finn Eggers Aug 24 '19 at 7:07
  • $\begingroup$ Ah, sorry about that. I had implicitly considered $0 \le k \le 1$. $\endgroup$ – Utkarsh Verma Aug 24 '19 at 9:20
1
$\begingroup$

Your upper bounding function is not correct. Choosing a $k>1$ would show a contradiction.

This problem can be approached by splitting into two cases: when $0\leq k \leq 1$ and when $k \geq 1$

When $k \geq 1$:

The upper bounding function can be attained by a binomial identity. Let $x_n = k^\frac{1}{n} - 1 \geq 0 $. Then we have:

$$ 1 + nx_n \leq (1+x_n)^n = k $$

Thus

$$k^\frac{1}{n} \leq 1 + \frac{k-1}{n} $$

The expression on the right converges to 1. A lower bounding function is obviously just 1.

When $0\leq k < 1$:

The upper bounding function is just 1.

The lower bounding function can be found using a similar technique as the first case, except now we use reciprocals. Let $x_n = k^{-\frac{1}{n}} - 1 \geq 0$.

Then we have:

$$1 + nx_n \leq (1 + x_n)^n = k^{-1}$$

Which results in:

\begin{align} k^{-\frac{1}{n}} - 1 &\leq \frac{1 - k}{nk} \\ k ^ \frac{1}{n} &\geq \frac{nk}{1 - k + nk} \end{align}

Since the right hand side tends to 1, we are done.

$\endgroup$
3
  • $\begingroup$ Nice answer. But being a beginner I'm still unable to fathom how you thought of using the binomial theorem just by seeing $k^\frac{1}{n}$. $\endgroup$ – Utkarsh Verma Aug 24 '19 at 9:24
  • 1
    $\begingroup$ This is a technique that I learnt from (currently) studying Rudin's Principles of Mathematical Analysis (Theorem 3.20(b) ). If it's of any comfort, I had to look up the technique before writing the answer. Its just one of those things that you would get better in time by being exposed to such techniques. $\endgroup$ – Sean Lee Aug 24 '19 at 15:01
  • $\begingroup$ If $k<1,$ then $1/k>1,$ which implies $(1/k)^{1/n} = 1/k^{1/n} \to 1$ by the $k>1$ case. This implies $k^{1/n}\to 1.$ $\endgroup$ – zhw. Aug 24 '19 at 18:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.