Find the limit of a real sequence $a_n = \dfrac{(n^2+n+1)^{10}-(n+1)^{20}}{(n^2+1)^{10}-(n+1)^{20}}$

Question

$$lim_{n\to\infty} \dfrac{(n^2+n+1)^{10}-(n+1)^{20}}{(n^2+1)^{10}-(n+1)^{20}} \quad (*)$$

My attempt at solution:

All my attempts lead to $\dfrac{0}{0}.$ For example:

$(*)\ = lim_{n\to\infty} \dfrac{(n^2+n+1)^{10}-(n^2+2n+1)^{10}}{(n^2+1)^{10}-(n^2+2n+1)^{10}} \\ = lim_{n\to\infty} \dfrac{n^{20}(1+\dfrac{1}{n}+\dfrac{1}{n^2})^{10}-n^{20}(1+\dfrac{2}{n}+\dfrac{1}{n^2})^{10}}{n^{20}(1+\dfrac{1}{n^2})^{10}-n^{20}(1+\dfrac{2}{n}+\dfrac{1}{n^2})^{10}} \\ = lim_{n\to\infty} \dfrac{(1+\dfrac{1}{n}+\dfrac{1}{n^2})^{10}-(1+\dfrac{2}{n}+\dfrac{1}{n^2})^{10}}{(1+\dfrac{1}{n^2})^{10}-(1+\dfrac{2}{n}+\dfrac{1}{n^2})^{10}} $

Any ideas please?

Answer 1

$$\lim_{n\to\infty} \dfrac{(n^2+n+1)^{10}-(n+1)^{20}}{(n^2+1)^{10}-(n+1)^{20}}=$$ $$=\lim_{n\rightarrow+\infty}\frac{(n^2+n+1-n^2-2n-1)((n^2+n+1)^9+...+(n^2+2n+1)^9)}{(n^2+1-n^2-2n-1)((n^2+1)^9+...+(n^2+2n+1)^9)}=\frac{-1\cdot10}{-2\cdot10}=\frac{1}{2}.$$

Answer 2

First, change variable $n = m-1$. That makes it look simpler: $$\frac{(m^2-2m+1)^{10}-m^{20}}{(m^2-2m+2)^{10}-m^{20}}$$

Then use the Binomial theorem to expand, and finally apply the method you were using.

$$\frac{-10m^{19} + ...\text{lower degree terms}}{-20m^{19}+...\text{lower degree terms}}\to\frac{1}{2}$$

Answer 3

$ \dfrac{(n^2+n+1)^{10}-(n+1)^{20}}{(n^2+1)^{10}-(n+1)^{20}}$

$\frac {n^{20} + 10n^{18}(n+1) + \text{less significant terms}) - (10^{20}+ 20*n^{19} + \text{less significant terms})}{(10^{20} + 10*n^{18} + \text{l.s.t.}) - (n^{20} + 20*n^{19} + 190*n^18 + \text{l.s.t})}$

$=\frac {-10*n^{19} + LST}{-20*n^{19} -180*n^{18}+ LST} \to\frac {-10*n^{19} + LST}{-20*n^{19} + LST}\to \frac 12$

...

Okay, a formal answer needs more formality but that is in essence it. Reduce to highest significant term.

Answer 4

Let us make it more general to see not only the limit but also how it is approached.

Making the same as you did, let us consider $$A_k=\dfrac{(1+\dfrac{1}{n}+\dfrac{1}{n^2})^{k}-(1+\dfrac{2}{n}+\dfrac{1}{n^2})^{k}}{(1+\dfrac{1}{n^2})^{k}-(1+\dfrac{2}{n}+\dfrac{1}{n^2})^{k}}$$ and, now, use the generalized binomial theorem or Taylor series remembering that $$(1+\epsilon)^k=1+k\epsilon+\frac 12k(k-1)\epsilon^2+O(\epsilon^3)$$ Using for the different pieces, we should get $$(1+\dfrac{1}{n}+\dfrac{1}{n^2})^{k}=1+\frac{k}{n}+\frac{k^2+k}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ $$(1+\dfrac{2}{n}+\dfrac{1}{n^2})^{k}=1+\frac{2 k}{n}+\frac{2 k^2-k}{n^2}+O\left(\frac{1}{n^3}\right)$$ $$(1+\dfrac{1}{n^2})^{k}=1+\frac{k}{n^2}+O\left(\frac{1}{n^3}\right)$$ Replacing in $A_k$, we then have $$A_k=\frac{-\frac k n - \frac{3 (k-1) k}{2n^2}+O\left(\frac{1}{n^3}\right)} {-\frac{2 k}{n}-\frac{2 (k-1) k}{n^2}+O\left(\frac{1}{n^3}\right) }=\frac{1}{2}+\frac{k-1}{4 n}+O\left(\frac{1}{n^2}\right)$$ So, as you can see, the limit does not depend on $k$.