What is the limit of $(3n^2 + n)^{1/n}$ and why?

Question

I am really unsure of how to answer this as I have tried taking out n and using the ratio lemma and am still confused!

Answer 1

Yes taking out $n$ is a good idea but $3n^2$ is a better one to obtain

$$(3n^2 + n)^{1/n}=(3n^2)^\frac1n\left(1 + \frac1{3n}\right)^{1/n}$$

and since

$$\left(1 + \frac1{3n}\right)^{1/n} \to 1^0=1$$

we reduce to evaluate the limit for $(3n^2)^\frac1n=3^\frac1n \,\left(n^\frac1n\right)^2$.

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As an alternative we can use that

$$1=1^\frac1n \le (3n^2 + n)^{1/n}\le (4n^2)^{1/n} =4^\frac1n \,\left(n^\frac1n\right)^2$$

Answer 2

$$L=\lim_{n \to \infty}(3n^2 + n)^{1/n}$$ Apply Cauchy-D'Alembert criterion: $$L=\lim_{n \to \infty}\dfrac {3n^2+7n+ 4}{3n^2+n}=\lim_{n \to \infty}\dfrac {3n^2}{3n^2}$$ $$\implies L=1$$

Answer 3

Try to factor $3n^2$ out.

$$(3n^2 + n)^{\frac{1}{n}} = (3n^2)^\frac{1}{n}(1 + \frac{1}{3n})^{\frac{1}{n}}$$

Now notice that

$$\lim_{n \rightarrow \infty} (3n^2)^\frac{1}{n} = 1$$ $$\lim_{n \rightarrow \infty} (1 + \frac{1}{3n})^{\frac{1}{n}} = 1$$

Hence

$$\lim_{n \rightarrow \infty} [(3n^2 + n)^{\frac{1}{n}}] = \lim_{n \rightarrow \infty} [(3n^2)^\frac{1}{n}(1 + \frac{1}{3n})^{\frac{1}{n}}] = 1 $$

Answer 4

$$L=\underset{n\to \infty }{\text{lim}}\left(3 n^2+n\right)^{1/n}$$ Take the logarithm of the limit $$\log L=\log \underset{n\to \infty }{\text{lim}}\left(3 n^2+n\right)^{1/n}=\underset{n\to \infty }{\text{lim}}\frac{\log \left(3 n^2+n\right)}{n}=0$$ If $\log L=0$ then $L=1$