Prove that $\lim_{n \to \infty} \sqrt[n]{a} = 1$ using only the basic properties of sequences

For $a\ge1$, $a^{1/n}\ge a^{1/(n+1)}\ge1$, so we have a decreasing sequence that is bounded below (by $1$). Therefore the limit exists, and we have $L=\lim_{n\to\infty}a^{1/n}\ge1$. But we also have

$$L=\lim_{n\to\infty}a^{2/2n}=(\lim_{n\to\infty}a^{1/2n})^2=L^2$$

so we must have $L=1$.

The same argument works for $a\le1$, except in this the inequalities are $a^{1/n}\le a^{1/(n+1)}\le1$, so we have an *increasing sequence that is bounded above (again by $1$).

It's easier to prove first for $a\geq 1$ and then use that to prove for $a<1$.

Specifically, we use that if $x>0$ then $(1+x)^n\geq 1+nx$. This is from the binomial theorem. If $x=\sqrt[n]a-1$ this means that $$a\geq 1+n\left(\sqrt[n]a -1\right),$$ and therefore that $0\leq\sqrt[n]{a}-1\leq \frac{a-1}{n}$.

The case $0<a<1$ is a little harder. You can use that $\sqrt[n]a=\frac1{\sqrt[n]{1/a}}$ and use the limit of $1/a$ to get the limit for $a$. Namely, if $b=\frac{1}{a}$:

$$0\leq 1-\sqrt[n]a=\frac{\sqrt[n]b -1}{\sqrt[n]{b}}\leq \sqrt[n]b-1$$

Another approach is to use that $x^n-1=(x-1)\left(1+x+x^2+\cdots+x^{n-1}\right)$. Then, with $x=\sqrt[n]a$ you get:

$$\sqrt[n]{a}-1=\frac{a-1}{\sqrt[n]{a}^{n-1}+\cdots+\sqrt[n]{a}+1}$$

When $a\geq 1$ this means that $0\leq \sqrt[n]{a}-1\leq \frac{a-1}{n}.$.

When $0<a<1$ you have $a\leq\sqrt[n]{a}^k$ for $k=1,\dots n-1$ so:

$$0\leq 1-\sqrt[n]{a}\leq \frac{1-a}{1+(n-1)a}$$

First, suppose $a>1$. For each $n\in\mathbb{N}$, let $a_n=\sqrt[n]{a}$. It's easy to see that $a_n >1$ for all $n$. Hence, for each $n$ we may write $a_n=1+d_n$, where $d_n>0$. We will show that $d_n\to 0$ as $n\to \infty$.

To that end, notice that

$$a=a_n^n=(1+d_n)^n\geq1+\binom{n}2d_n^2,$$

where the inequality comes from the binomial theorem. Then:

$$d_n^2=\frac{2(a-1)}{n(n-1)}$$

The numerator is a positive constant, and the denominator goes to infinity as $n\to\infty$. it follows $d_n^2\to0$ as $n\to\infty$, and so too does $d_n\to 0$ as $n\to\infty$.

Using this approach, do you think you can tackle the $0<a<1$ case?

Suppose $a>1$ then $\sqrt[n]{a}>1$

for any $\varepsilon>0$ we will find an index $p$ such that for $n>p$ it happens that $\sqrt[n]{a}-1<\varepsilon$

$$a^{1/n}<\varepsilon+1 \to \frac{\log a}{n}<\log(\varepsilon+1) \to \frac{n}{\log a}>\frac{1}{\log(\varepsilon +1)}\to n>\frac{\log a}{\log(\varepsilon+1)}$$ So taking $p=\left\lceil \dfrac{\log a}{\log(\varepsilon+1)}\right\rceil$ we are sure that for $n>p$ we have $\sqrt[n]{a}-1<\varepsilon$

For instance consider $\sqrt[n]{2}\to 1$ as $n\to\infty$

take $\varepsilon=\dfrac{1}{10000}$ and $p=\left\lceil \dfrac{\log 2}{\log(\frac{1}{10000}+1)}\right\rceil=6932$

Indeed for $n\ge 6932$ let's say $n=8000$ we have $\sqrt[8000]{2}<1+\frac{1}{10000}=1.0001$. Actually $\sqrt[8000]{2}\approx 1.0000866<1.0001$

Hope this helps

Edit

In a similar way can be done for $0<a<1$