Proof Verification: $\lim{C^{1/n}}=1$ for $C>0$

Question

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?

Answer 1

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 $C<1$.

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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?

Answer 2

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.