Limit of $(3n+1)^{1/n}$

Question

I am working through a question that requires me to prove that a certain limit tends to 1. I have reached an endpoint where I have:

$$\lim\limits_{n\to \infty} (3n+1)^{1/n}$$

I know that and can use the fact that $\lim\limits_{n\to \infty}( n^{1/n}) = 1$

However, I am not sure if I can just use this fact to say that it applies for $(3n+1)^{1/n} $ and I also am stuck on proving it.

Answer 1

Answer:

$(3n+1)^{\frac{1}{n}} =e^{\frac{1}{n}ln(3n+1)}=e^{\frac{1}{n}ln(n(3+\frac{1}{n} )) }=e^{\frac{1}{n}ln(n) +\frac{1} {n} ln(3+\frac{1}{n} )}$

So :

$\lim_{n\to+\infty} (3n+1)^{\frac{1}{n}}=\lim_{n\to+\infty} e^{\frac{1}{n}ln(n) +\frac{1} {n} ln(3+\frac{1}{n} )}=1$

Because

$\lim_{n\to+\infty} \frac{ln(n)}{n} =0$

And

$\lim_{n\to+\infty}\frac{1} {n}ln(3+\frac{1}{n} ) =0$

Answer 2

As a hint you apply $\ln$ to both sides of limit.

As a second hint rewrite like below $$ \lim_{n\to \infty} (3n+1)^{1/n}=\\\lim_{n\to \infty}\sqrt[n]{3n+1}=\lim_{n\to \infty}\sqrt[n]{3}\sqrt[n]{n+\frac13}=\\ \lim_{n\to \infty} \sqrt[n]{3}\lim_{n\to \infty}\sqrt[n]{n+\frac13}=1\times1$$

Answer 3

Per the hint in comments, we squeeze the limit expression between $(3n)^{1/n}$ and $(4n)^{1/n}$, both of which evaluate to $1$, so the limit is $1$. (It's easier to show that $\lim_{n\to\infty}a^{1/n}=1$ for $a\ge1$.)

Answer 4

We can use that

$$1=(1)^{1/n}\le (3n+1)^{1/n}\le (4n)^{1/n}=4^\frac1n \cdot n^\frac1n$$

Answer 5

$$L=\lim_{n\to \infty} (3n+1)^{1/n}$$ Apply D'Alembert-Cauchy criterion: $$L=\lim_{n\to \infty} \dfrac {3n+4}{3n+1}$$ $$\implies L=1$$

Answer 6

\begin{align*} \lim_{n \rightarrow \infty} (3n+1)^{1/n} &= \lim_{n \rightarrow \infty} \exp \left( \ln \left( (3n+1)^{1/n} \right) \right) \\ &= \lim_{n \rightarrow \infty} \exp \left( \frac{1}{n} \ln \left( 3n+1 \right) \right) \\ &= \exp \left( \lim_{n \rightarrow \infty} \frac{\ln \left( 3n+1 \right)}{n} \right) \\ &= \exp \left( \lim_{n \rightarrow \infty} \frac{\ln \left( 3n \left( 1 + \frac{1}{3n} \right) \right)}{n} \right) \\ &= \exp \left( \lim_{n \rightarrow \infty} \frac{\ln (3) + \ln (n) + \ln \left( 1 + \frac{1}{3n} \right)}{n} \right) \\ &\overset{?_1}= \exp \left( \lim_{n \rightarrow \infty} \frac{\ln (3)}{n} + \lim_{n \rightarrow \infty} \frac{\ln (n)}{n} + \lim_{n \rightarrow \infty} \frac{\ln \left( 1 + \frac{1}{3n} \right)}{n} \right) \\ &= \exp \left( 0 + \lim_{n \rightarrow \infty} \ln n^{1/n} + \lim_{n \rightarrow \infty} \frac{\ln \left( 1 + \frac{1}{3n} \right)}{n} \right) \\ &\overset{?_2}= \exp \left( \ln \lim_{n \rightarrow \infty} n^{1/n} + \lim_{n \rightarrow \infty} \frac{\ln \left( 1 + \frac{1}{3n} \right)}{n} \right) \\ &= \exp \left( \ln 1 + \lim_{n \rightarrow \infty} \frac{\ln \left( 1 + \frac{1}{3n} \right)}{n} \right) \\ &\overset{?_3}= \exp \left( 0 + \frac{\lim_{n \rightarrow \infty} \ln \left( 1 + \frac{1}{3n} \right)}{n} \right) \\ &\overset{?_4}= \exp \left( \frac{ \ln \lim_{n \rightarrow \infty} \left( 1 + \frac{1}{3n} \right)}{n} \right) \\ &\overset{?_5}= \exp \left( \frac{ \ln \left( \lim_{n \rightarrow \infty} 1 + \lim_{n \rightarrow \infty} \frac{1}{3n} \right) }{n} \right) \\ &= \exp \left( \frac{ \ln \left( 1 + 0 \right) }{n} \right) \\ &= \exp \left( \frac{ \ln 1 }{n} \right) \\ &= \exp \left( \frac{ 0 }{n} \right) \\ &= \exp \left(0 \right) \\ &= 1 \text{.} \end{align*}

We use the continuity of the exponential on its domain without further comment.

• $?_1$ : The limit of a sum is a sum of the limits if those limits exist. We subsequently show that they do, so this equality is valid.
• $?_2$ : The logarithm is continuous on its domain, so this equality holds if the new limit is in $(0,\infty)$. It is $1$, so this equality is valid.
• $?_3$ : Again w exchange the limit and the logarithm. Again, the new limit is in $(0,\infty)$, so this equality is valid.
• $?_4$ : Division is continuous in its numerator without any restriction on domain, so this equality is valid. (Notice that the domain of the denominator is restricted, so there can be something to check here.)
• $?_5$ : The limit of a sum is the sum of the limits, if those limits exist. As we subsequently show, those limits do exist, so this equality is valid.