I have the following task:
Justify the following statements without using continuity of elementary functions (i.e. it has not yet been proved that $x_{\{n\}} \to a$, as $n \to \infty$ implies that $f(x_{\{n\}}) \to f(a)$, as $n \to \infty$):
$$\lim_{k\to\infty} \sqrt[k]{a}=1, a > 0 $$ $$\lim_{k\to\infty} \sqrt[k]{k}=1$$ $$\lim_{k\to\infty} \frac{k^\alpha}{b^k}=0, b > 1$$
I tried to use defenition of limit $|x_{n} - A| < \xi$, but I am not sure in this approach.
For $a>1$ $$a=[1+(\sqrt[n]{a}-1)]^n > n(\sqrt[n]{a}-1)$$ and we obtain $0<\sqrt[n]{a}-1<\frac{a}{n}<\varepsilon$, when $n>\frac{a}{\varepsilon}$.
For $0<a<1$ will be $b=\frac{1}{a}>1$ and we can use already obtained result, or directly $|\sqrt[k]{a}-1|=\left| \frac{1}{\sqrt[k]{b}} - 1 \right|<|\sqrt[k]{b}-1|$.
For second example works same approach.
For third, when $\alpha \in \mathbb{N}$ is enough to consider $\frac{k}{b^k}$, for $b>1$: $$\frac{k}{b^k}=\frac{k}{[1+(b-1)]^k}=\frac{k}{1+(b-1)k+\frac{k(k-1)}{2}(b-1)^2+\cdots +(b-1)^k}<\frac{2k}{k(k-1)(b-1)^2}=\\=\frac{2}{(k-1)(b-1)^2}$$
Addition. Let's consider any $\varepsilon >0$. $$\frac{2}{(k-1)(b-1)^2}< \varepsilon \Leftrightarrow \frac{2}{\varepsilon(b-1)^2}< (k-1)$$ Now if we choose $K=\left\lfloor \frac{2}{\varepsilon(b-1)^2} +1\right\rfloor+1 \in \mathbb{N}$, then for $k>K$ holds $\frac{k}{b^k}< \varepsilon$.