How do you prove $\lim_{n\to\infty} 1/n^{1/n}$ using only basic limit theorems? How do you prove $\lim_{n\to\infty} \frac {1}{n^{1/n}}$ using only basic limit theorems? I thought it was $0$, but my book lists the solution as $1$. How come?
 A: Consider $$a_n=\frac {1}{n^{1/n}}$$ and take logarithms $$\log(a_n)=\frac 1 n \,\log(\frac 1 n)=-\frac{\log(n)}n$$ So, since $n$ varies faster then $\log(n)$, then $\log(a_n)\to 0$ and so $a_n\to 1$.
A: Hint: Use $ \lim_{n\to \infty} (a_n)^{1/n}=l $ if $\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=l$
A: Method $1$: Clearly, $n^{1/n}>1$ for all $n \in \mathbb{N}$. Let $n^{1/n} = (1+a)$. We will prove that $a \to 0$ as $n \to \infty$. For $n \geq 2$, we have
$$n = (1+a)^n \geq \dfrac{n(n-1)}2a^2 \implies a^2 \leq \dfrac2{n-1}$$
Hence, as $n \to \infty$, we have $a \to 0$, which means that $n^{1/n} \to 1$.

Method $2$: From AM-GM, for $a > 1$, we have
$$\dfrac{a+a+\cdots+a+n}n \geq \sqrt[n]{na^{n-1}} \implies (1+a) - \dfrac{a}n \geq n^{1/n}a^{1-1/n}$$
Taking the limit as $n \to \infty$, we obtain
$$\lim_{n \to \infty} n^{1/n}a \leq (1+a) \implies \lim_{n \to \infty} n^{1/n} \leq 1+\dfrac1a$$
for all $a \in \mathbb{R}$. Taking the limit as $a \to \infty$, we obtain
$$n^{1/n} \leq 1$$
Further, $n^{1/n} \geq 1$ for all $n$. Hence, we have $\lim_{n \to \infty} n^{1/n}=1$.
A: $$\lim_{n\to\infty}\frac{1}{n^{\frac{1}{n}}}=\lim_{n\to\infty}n^{-\frac{1}{n}}=\lim_{n\to\infty}\exp\left(\ln\left(n^{-\frac{1}{n}}\right)\right)=\lim_{n\to\infty}\exp\left(-\frac{1}{n}\ln\left(n\right)\right)=$$
$$\lim_{n\to\infty}\exp\left(-\frac{\ln\left(n\right)}{n}\right)=\exp\left(-\lim_{n\to\infty}\frac{\ln\left(n\right)}{n}\right)=\exp\left(-\lim_{n\to\infty}\frac{\frac{\text{d}}{\text{d}n}\left(\ln\left(n\right)\right)}{\frac{\text{d}}{\text{d}n}\left(n\right)}\right)=$$
$$\exp\left(-\lim_{n\to\infty}\frac{\frac{1}{x}}{1}\right)=\exp\left(-\lim_{n\to\infty}\frac{1}{x}\right)=\exp\left(-0\right)=\exp(0)=e^0=1$$
