Show that $\lim_{n\to \infty} \left(\frac1n\right)^{\frac1n}=1$

Question

Can't figure out how to show

$$\lim_{n\to \infty} \left(\frac1n\right)^{\frac1n}=1$$

Any help?

Answer 1

Hint: $\lim_{x\to 0^+}x^x=1$ by

$$\ln\left(\lim_{x\to 0^+}x^x\right)=\lim_{x\to 0^+} (x\ln x)=\lim_{x\to 0^+}\frac{\ln x}{\frac{1}{x}}\stackrel{\text{L'Hop}}=\lim_{x\to 0^+}\frac{\frac{1}{x}}{-\frac{1}{x}\frac{1}{x}}=\lim_{x\to 0^+}(-x)=0$$

How is this fact related to your limit? I assume you want $\lim_{n\to\infty}(1/n)^{1/n}$.

Answer 2

First note that $n^{\frac{1}{n}}\geq 1$, for all $n \geq 1$.

Let $n^{\frac{1}{n}}= 1+ x_n$; where $x_n\geq 0$. Suffices to show $x_n \to 0$ as $n$ tends to $\infty$ to get the desired limit as 1. Note that $ (1+x_n)^n = n$. Now using binomial expansion, we have $\frac{n(n-1)}{2}{x_n}^2 \leq n,$ for all $n \geq 2$.

This gives us ${x_n}^2 \leq \frac{2}{n-1}$, for all $n\geq 2$. Hence $x_n \to 0$, as $n$ tends to $\infty$.

Answer 3

I will assume that you are referring to $\lim_{n\rightarrow\infty}(\frac{1}{n})^\frac{1}{n}$ because $\lim_{n\rightarrow 0}(\frac{1}{n})^\frac{1}{n}$ diverges. what you need to do to solve this is to take the natural log of the function, giving you $\lim_{n\rightarrow \infty}\frac{\log(n)}{n}$ and because $n$ grows a whole lot faster than $\log(n)$ this clearly approaches $0$. Note that now we have not found $\lim_{n\rightarrow\infty}(\frac{1}{n})^\frac{1}{n}$, we have found $\lim_{n\rightarrow\infty}\log((\frac{1}{n})^\frac{1}{n})$ to convert this back to our original limit we can simply find $\lim_{n\rightarrow\infty}e^{(\log((\frac{1}{n})^\frac{1}{n}))} = e^{\lim_{n\rightarrow\infty}\log((\frac{1}{n})^\frac{1}{n})}$ and from earlier, we know that this whole exponent is simply $0$, so we have $e^0=1$.

Answer 4

Note that for positive $x$ we have $0 \le x e^{-x} = {x \over {e^x}} \le {x \over 1+x + {1 \over 2} x^2}$, and so we have $\lim_{x \to \infty} x e^{-x} = 0$.

Let $x_n = - \ln {1 \over n}$, and note that $\lim_{n \to \infty} x_n = \infty$, and so $\lim_{n \to \infty} x_n e^{-x_n} = \lim_{n \to \infty} (- {1 \over n} \ln {1 \over n}) = \lim_{n \to \infty} (- \ln({1 \over n})^{1 \over n}) = 0$.

Consequently, since $\exp$ is continuous at $x=0$, we have $\lim_{n \to \infty} e^{\ln({1 \over n})^{1 \over n}} = \lim_{n \to \infty} ({1 \over n})^{1 \over n} = 1$.