How to evaluate $\lim_{n\to\infty}k^{\frac{1}{n}}$ using the Sandwich theorem?

Question

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.

Answer 1

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.