How can I deal with radicals when I need to calculate limits?

Question

I don't know what to do with radicals in limits. For example, is it obvious that $ \lim_{n\to \infty}\left(\sqrt[n]{c}\right)=1$ where c is a constant? I am facing with something extremely hard for me. Like that: $$\lim_{n\to \infty}\left( \frac{\sqrt[n]{n^3}+\sqrt[n]{7}}{3\sqrt[n]{n^2}+\sqrt[n]{3n}} \right)$$ L'Hôpital's rule is prohibited here.

Answer 1

Hint: Apply the following limit.

Proof that $\boldsymbol{\lim\limits_{n\to\infty}n^{\frac1n}=1}$

For $n\ge e$, $$ \begin{align} \frac{(n+1)^{\frac1{n+1}}}{n^{\frac1n}} &=\frac{\left(1+\frac1n\right)^{\frac1{n+1}}}{n^{\frac1{n(n+1)}}}\tag{1}\\ &\le\frac{1+\frac1{n(n+1)}}{n^{\frac1{n(n+1)}}}\tag{2}\\[9pt] &\le1\tag{3} \end{align} $$ Explanation
$(1)$: divide numerator and denominator by $n^{\frac1{n+1}}$
$(2)$: Bernoulli's Inequality
$(3)$: $e^x\ge1+x$

Thus, for $n\ge3$, $n^{\frac1n}$ is a decreasing function bounded below by $1$. Therefore, $a=\lim\limits_{n\to\infty}n^{\frac1n}$ exists and is not less than $1$.

$$ \begin{align} a &=\lim_{n\to\infty}(2n)^{\frac1{2n}}\tag{4}\\ &=\lim_{n\to\infty}2^{\frac1{2n}}\left(\lim_{n\to\infty}n^{\frac1n}\right)^{\frac12}\tag{5}\\[6pt] &=1\cdot a^{\frac12}\tag{6}\\[12pt] &=1\tag{7} \end{align} $$ Explanation:
$(4)$: limit of every other term is the limit of every term
$(5)$: product of limits is the limit of the product
$(6)$: Bernoulli's Inequality says $2^{\frac1{2n}}=(1+1)^{\frac1{2n}}\le1+\frac1{2n}$
$(7)$: multiply the reciprocal of the left side of $(4)$ by the square of $(6)$

QED

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