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.

  • $\begingroup$ 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

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:

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


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

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.

  • $\begingroup$ 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
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    $\begingroup$ 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

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