Proof Verification: $\lim{C^{1/n}}=1$ for $C>0$ Claim: $\lim{C^{1/n}}=1$ for $C>0$
Working: 
$|C^{1/n}-1| < \epsilon$
$ \implies C^{1/n} < \epsilon +1$
$\implies {(1/n)}\ln{C} < \ln{(\epsilon +1)}$
$ \implies n> {\ln{C}/\ln{(\epsilon +1)}}$
Proof:
Let $ \epsilon >0$ be given.
Choose $N>{\ln{C}/\ln{(\epsilon +1)}}$
Then for any $n>N$,  this implies that  $|C^{1/n} -1|<\epsilon$
Hence, $\lim{C^{1/n}} =1$ for all $C>0$
Can anyone please verify this proof?
 A: It is not correct.
For one, you did away with the absolute value.
Try to split it into cases. $C=1$ is obvious, but consider $C>1$. Afterwards, a simple trick takes care of the case $C<1$.

If you want to avoiding logarithms, consider the following.
Let $C>1$, so that $C^{1/n}>1$ for all $n\geq 1$. For each $n$, write $C^{1/n}= 1+ d_n$ and notice $d_n > 0$.
Then, for each $n$, we have by Bernoulli's ineuqality that
$$C = {\left(C^{1/n}\right)}^n =(1+d_n)^n \geq 1+nd_n.$$
This implies that $0<d_n \leq \frac{C-1}{n}$ and hence $\lim_{n\to\infty}d_n = 0$ by the squeeze theoerem.
It follows that
$$\lim_{n\to\infty} C^{1/n} = \lim_{n\to\infty} 1+ d_n = 1,$$
as desired. Finally, for $0<C<1$, we have
$$C^{1/n} = \frac{1}{\left(\frac1C\right)^{1/n}}.$$
Do you think you can conclude?
A: Here is a simpler proof without having to consider different cases (other than implicitly):
\begin{align}
\bigl|C^{1/n}-1\bigr|<\varepsilon&\iff 1-ε< C^{1/n} < 1 +ε\iff \ln(1-ε)<\frac 1n\ln C <\ln(1+ε)\\
&\iff (n\ln(1-ε)\color{red}< \ln C)\quad\text{and}\quad (\ln C<n\ln(1+ε))\\
&\iff n\color{red}>\underbrace{\frac{\ln C}{\ln(1-ε)}}_{\text{because }\ln(1-ε)<0}\quad\text{and}\quad n>\frac{\ln C}{\ln(1+ε)}\\
&\iff n>\max\biggl(\frac{\ln C}{\ln(1-ε)},\frac{\ln C}{\ln(1+ε)}\biggr).
\end{align}
Note  that if $C\ne 1$, one of these values is positive, the other is negative.
