Calculate the limit of a series number

Question

I have a homework, about calculate the limit of a series: $$ \lim\limits_{n \to +\infty} \dfrac{\sqrt[n] {n^3} + \sqrt[n] {7}}{3\sqrt[n]{n^2} + \sqrt[n]{3n}} $$ Solution is $\frac{1}{2}$. I am trying use the unequality: $$ \dfrac{\sqrt[n] {n^3} }{3\sqrt[n]{n^2} + \sqrt[n]{3n}} \le \dfrac{\sqrt[n] {n^3} + \sqrt[n] {7}}{3\sqrt[n]{n^2} + \sqrt[n]{3n}} \le \dfrac{\sqrt[n] {n^3} + \sqrt[n] {7}}{3\sqrt[n]{n^2}} $$ However, I haven't got to find solution.

Answer 1

It is true that $\lim_{n\to\infty} \sqrt[n]{n} \to 1$ and $\lim_{n\to\infty} \sqrt[n]{i} \to 1$ for some natural number $i$.

Rearranging, we see that $\sqrt[n]{n^2} = \sqrt[n]{n}\cdot\sqrt[n]{n}$

Thus $\lim_{n\to\infty} \sqrt[n]{n^2} = 1$

Applying similar rules and the algebra of limits we should see that

$$\lim_{n\to\infty} \frac{\sqrt[n]{n^3} + \sqrt[n]{7}}{3\sqrt[n]{n^2} + \sqrt[n]{3n}} = \frac{1 + 1}{3\cdot1 + 1} = \frac{1}{2} $$

An aside, It's every important to know that if $\lim_{n\to n_0} f(n) = A$ and $\lim_{n\to n_0} g(n) = B$ then $\lim_{n \to n_0} f(n) + g(n) = A + B$. Same goes for fractions (of course barring B=0).

Answer 2

I'd calculate the individual limits then put them together.

$\lim\limits_{n \to +\infty} \sqrt[n] {n^3} = \lim\limits_{n \to +\infty} \exp(ln(\sqrt[n] {n^3})) = \lim\limits_{n \to +\infty} \exp(ln(({n^3})^\dfrac{1}{n})) $

Appyling log properties: $log(A^n) = n*log(A)$

$\lim\limits_{n \to +\infty} \exp(\dfrac{1}{n} * ln( {n^3})) = \lim\limits_{n \to +\infty} \exp(\dfrac{ln( {n^3})}{n}) = \lim\limits_{n \to +\infty} \exp(\dfrac{3*ln( {n})}{n})$

Both the numerator's and the denomitator's limits goes to $+\infty$, so we can apply L'Hopital's rule:

$\lim\limits_{n \to +\infty} \exp(\dfrac{\dfrac{3}{n}}{1}) = \lim\limits_{n \to +\infty} \exp(\dfrac{3}{n}) = \exp(0) = 1 $

The limits remaining are pretty similar, so I will be more straightfoward.

$(*)$ means I used L'Hopital's rule:

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$\lim\limits_{n \to +\infty} \sqrt[n] {7} = \lim\limits_{n \to +\infty} \exp(\dfrac{ln(7)}{n}) = \exp(0) = 1$

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$\lim\limits_{n \to +\infty} 3\sqrt[n] {n^2} = 3\lim\limits_{n \to +\infty} \exp(ln(\sqrt[n] {n^2})) = 3\lim\limits_{n \to +\infty} \exp(\dfrac{2ln(n)}{n}) =(*) 3\lim\limits_{n \to +\infty} \exp(\dfrac{\dfrac{2}{n}}{1}) = 3 $

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$\lim\limits_{n \to +\infty} \sqrt[n] {3n} = \lim\limits_{n \to +\infty} \exp(ln(\sqrt[n] {3n})) = \lim\limits_{n \to +\infty} \exp(\dfrac{ln(3n)}{n}) =(*)\lim\limits_{n \to +\infty} \exp(\dfrac{\dfrac{1}{n}}{1}) = 1 $

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Putting it all together:

$ \lim\limits_{n \to +\infty} \dfrac{\sqrt[n] {n^3} + \sqrt[n] {7}}{3\sqrt[n]{n^2} + \sqrt[n]{3n}} = \dfrac{1+1}{3+1} = \dfrac{1}{2}$

Using $ x = \exp(ln(x)) $ is very common for exponential limits.