Method to find $\lim_{n\to \infty}\frac{\sqrt[5]{n^3 + 3 \sqrt[5]{n^2}+1} - \sqrt[5]{n^3 + 2 \sqrt[5]{n}+1}}{1-\cos(\frac{3}{n})}$

Question

I have the following limit:

$$\displaystyle \lim_{n\to \infty}\frac{\sqrt[5]{n^3 + 3 \sqrt[5]{n^2}+1} - \sqrt[5]{n^3 + 2 \sqrt[5]{n}+1}}{1-\cos\left(\dfrac{3}{n}\right)}$$

I know that it should be $\dfrac{2}{15},$ however I could not find the method to solve it.

I had a hint to use Heine, so I think it can be rewritten as the limit of the function but I could not see how it can help me. I tried L'Hôpital's rule and also used $A^5 - B^5 = (A - B)(B^4 + B^3A + B^2A^2 + BA^3 + A^4),$ but nothing seems to help. I think I am missing some trick using standard (known) limits.

Answer 1

Since $$2+\frac{2}{5}=\frac{3\cdot4}{5},$$ by your work we obtain: $$\frac{\sqrt[5]{n^3 + 3 \sqrt[5]{n^2}+1} - \sqrt[5]{n^3 + 2 \sqrt[5]{n}+1}}{1-\cos(\frac{3}{n})}=$$ $$=\frac{3\sqrt[5]{n^2}-2\sqrt[5]{n}}{2\sin^2\frac{3}{2n}\left(\sqrt[5]{\left(n^3 + 3 \sqrt[5]{n^2}+1\right)^4} +...+ \sqrt[5]{\left(n^3 + 2 \sqrt[5]{n}+1\right)^4}\right)}=$$ $$=\frac{n^2\left(3\sqrt[5]{n^2}-2\sqrt[5]{n}\right)}{2\cdot\frac{\sin^2\frac{3}{2n}}{\frac{1}{n^2}}\cdot n^{\frac{3\cdot4}{5}}\left(\sqrt[5]{\left(1 + \frac{3 \sqrt[5]{n^2}}{n^3}+\frac{1}{n^3}\right)^4} +...+ \sqrt[5]{\left(1 + \frac{2 \sqrt[5]{n}}{n^3}+\frac{1}{n^3}\right)^4}\right)}\rightarrow\frac{3}{2\cdot\frac{9}{4}\cdot5}=\frac{2}{15}.$$ I used $$\frac{\sin^2\frac{3}{2n}}{\frac{1}{n^2}}=\frac{9}{4}\left(\frac{\sin\frac{3}{2n}}{\frac{3}{2n}}\right)^2\rightarrow\frac{9}{4}$$ because $$\lim_{x\rightarrow0}\frac{\sin{x}}{x}=1.$$

Answer 2

Hint: By the MVT, if $0<x<y,$ then

$$y^{1/5}-x^{1/5} = (1/5)c^{-4/5}(y-x).$$

The expression on the right is bounded below by $(1/5)y^{-4/5}(y-x)$ and bounded above by $(1/5)x^{-4/5}(y-x).$ And of course $1-\cos u \approx u^2/2$ as $u\to 0.$

Answer 3

Substitute $n=1/t$; if the limit exists for $t\to0$, then it's equal to the limit of the sequence. Then observe that $$ \sqrt[5]{\frac{1}{t^3}+\frac{3}{t^{2/5}}+1} =\frac{1}{t^{3/5}}\sqrt[5]{1+3t^{13/5}+t^3} =\frac{1}{t^{3/5}}\left(1+\frac{3}{5}t^{13/5}+o(t^{13/5})\right) $$ Similarly, $$ \sqrt[5]{\frac{1}{t^3}+\frac{3}{t^{1/5}}+1} =\frac{1}{t^{3/5}}\sqrt[5]{1+3t^{14/5}+t^3} =\frac{1}{t^{3/5}}(1+o(t^{13/5})) $$ Therefore the numerator is $$ \frac{1}{t^{3/5}}\left(\frac{3}{5}t^{13/5}+o(t^{13/5})\right)=\frac{3}{5}t^2+o(t^2) $$ The denominator is $$ 1-\cos3t=\frac{9t^2}{2}+o(t^2) $$ and therefore the limit is $$ \frac{3}{5}\frac{2}{9}=\frac{2}{15} $$

Answer 4

You could also make use of the fundamental limit $$\frac{(1+\alpha(x))^\gamma-1}{\alpha(x)} \to \gamma,$$ toghether with, of course, $$\frac{1-\cos\alpha(x)}{\alpha^2(x)}\to \frac12$$ whenever $\alpha(x)\to 0$. So \begin{eqnarray} \mathcal L &=&\lim_{n\to \infty}\frac{\sqrt[5]{n^3+3\sqrt[5]{n^2}+1}-\sqrt[5]{n^3+2\sqrt[5]{n}+1}}{1-\cos\frac3n}=\\ &\stackrel{\sqrt[5]{n}=t}{=}&\lim_{t\to \infty}\frac{\sqrt[5]{t^{15}+3t^2+1}-\sqrt[5]{t^{15}+2t+1}}{1-\cos\frac{3}{t^5}}=\\ &=&\lim_{t\to \infty}\underbrace{\frac{\frac9{t^{10}}}{1-\cos \frac3{t^5}}}_{\to 2}\cdot\frac{t^{10}}9\cdot t^3 \left[\left(\sqrt[5]{1+\frac{3t^2+1}{t^{15}}}-1\right)-\left(\sqrt[5]{1+\frac{2t+1}{t^{15}}}-1\right)\right]=\\ &=&\lim_{t\to \infty}\frac29\cdot t^{13}\left(\underbrace{\frac{\sqrt[5]{1+\frac{3t^2+1}{t^{15}}}-1}{\frac{3t^2+1}{t^{15}}}}_{\to\frac15}\cdot \frac{3t^2+1}{t^{15}}-\underbrace{\frac{\sqrt[5]{1+\frac{2t+1}{t^{15}}}-1}{\frac{2t+1}{t^{15}}}}_{\to \frac15}\cdot \frac{2t+1}{t^{15}}\right)=\\ &=&\lim_{t\to \infty}\frac2{45}\frac{3t^{15}-2t^{14}}{t^{15}}=\frac2{15}. \end{eqnarray}