Show $\sqrt[n]{b_n}\to 1$, given $\sqrt[n]{a_n} \to 1$ and $\frac{b_n}{a_n} \to g\in (0, \infty)$ Suppose that $\lim_{n\rightarrow\infty}\sqrt[n]{a_n}=1$ and $\lim_{n\rightarrow\infty}\frac{b_n}{a_n}=g$ where $g\in (0,\infty)$. We must prove that $\lim_{n\rightarrow\infty} \sqrt[n]{b_n}=1$.
Intuition tells me that this should be done by contradiction. But I can't get it to work. Here is what I came up with so far:
We know that for every $\epsilon>0$, for sufficiently large $n$ we have $\sqrt[n]{a_n}<\epsilon+1$. Suppose that $\lim_{n\rightarrow\infty} \sqrt[n]{b_n}\neq 1$. That means that there exists $\epsilon'$ such that, for sufficiently large $n'$ we have $\sqrt[n]{b_n}>\epsilon' + 1$.
Let's take $\epsilon'$ and $n=\max(n, n')$. Then we have:
$\sqrt[n]{b_n}>\epsilon' + 1$
$\sqrt[n]{a_n}<\epsilon' + 1$
And here, I think, something smart should be done to conclude that, in that case $\lim_{n\rightarrow\infty}\frac{b_n}{a_n}\neq g$. But I just can't see what to do. Any hints?
 A: Assume $\lim_{n \to \infty} (b_n)^{1/n} = M \ne 1$  and assume that the limit of $b_n$ exists:
$|(b_n)^{1/n} - M| < \epsilon$ if $n > N$ 
Also, $\lim_{n\to 1} (b_n)^{1/n} = b_1 \implies | (b_n)^{1/n} - 1| < \epsilon$
Let $\epsilon = |b_1 - M|/2$ 
$|b_1 - M|/2 = |b_1 - (b_n)^{1/n} + (b_n)^{1/n} - M|/2 = |-((b_n)^{1/n} - b_1) + ((b_n)^{1/n} - M)|/2 < |(b_n)^{1/n} - b_1|/2 + |(b_n)^{1/n} - M|/2$ 
$\therefore |(b_n)^{1/n} - M| < |b_1 - M|/2 < |(b_n)^{1/n} - M|/2 < |(b_n)^{1/n} - b_1|/2 + |(b_n)^{1/n} - M|/2 < |b_1 - M|$
A contradiction.
A: Is this solution correct?
For sufficiently large $n$:
$\frac{g}{2} < \frac{b_n}{a_n} < 2g$. If we take the n-th degree root of all these terms we get:
$\sqrt[n]{\frac{g}{2}} < \frac{\sqrt[n]{b_n}}{\sqrt[n]{a_n}}<\sqrt[n]{2g}$. $g$ is finite so $\lim$ of left and right side evaluates to $1$. That means that $\lim  \frac{\sqrt[n]{b_n}}{\sqrt[n]{a_n}}=1$. But $\lim \sqrt[n]{a_n}=1$ so in order for that to be true $\lim \sqrt[n]{b_n}$ must be equal to $1$ as well.
A: Let $\varepsilon > 0$. From $\lim_{n \to \infty} b_n/a_n = g \in (0,\infty)$ follows that there exists $n_0 \in \mathbb{N}$ s.t.
\begin{align*}
n \geq n_0 &\implies \left| \frac{b_n}{a_n} - g \right| < \varepsilon \\
&\iff g - \varepsilon < \frac{b_n}{a_n} < g + \varepsilon
\end{align*}
We can assume that $g - \varepsilon > 0$. Now 
\begin{align*}
\implies \sqrt[n]{g-\varepsilon} &< \frac{\sqrt[n]{b_n}}{\sqrt[n]{a_n}} < \sqrt[n]{g + \varepsilon} \\
\iff \sqrt[n]{g-\varepsilon}\sqrt[n]{a_n} &< \sqrt[n]{b_n} < \sqrt[n]{g + \varepsilon}\sqrt[n]{a_n}
\end{align*}
The left and right sides of the last inequality tend to $1$ as $n \to \infty$, so the result follows.
