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

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.

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

• Thank you. Now I am not so scared. But what about $\lim_{n\to \infty}(\sqrt[n]{c})=1$ where $c$ is a constant. Should I prove it? Is it obvious? – Okumo Dec 11 '16 at 20:07
• Look at step $(6)$ above. For $c\ge1$, we can use the same idea to get \begin{align} c^{\frac1n} &=\left(1+(c-1)\right)^{\frac1n}\\ &\le1+\frac{c-1}n \end{align} with Bernoulli's Identity. Alternatively, if $n^{\frac1n}\to1$, $c$ will eventually be less than $n$ at some point, so $$1\le\lim_{n\to\infty}c^{\frac1n}\le\lim_{n\to\infty}n^{\frac1n}=1$$ – robjohn Dec 11 '16 at 21:27
• Hmm, okay. But why can't I prove it easier? $$\lim_{n\to \infty}(\sqrt[n]{c})=\lim_{n\to \infty}(c^{1/n})=\lim_{n\to \infty}(c^{0})=1$$ – Okumo Dec 11 '16 at 23:13
• That works, but often people are not comfortable moving the limit to the exponent. I figured if you were comfortable with that, you wouldn't have asked about it. – robjohn Dec 11 '16 at 23:51
• I realize it only now :) Thank you very much. – Okumo Dec 12 '16 at 0:05