Find the limit $\lim_{n\to\infty}\left(\frac{n^2-4n+3}{n^2-7n+10}\right)^{n+\sin(n!)}$

Question

Find the limit $$\lim_{n\to\infty}\left(\dfrac{n^2-4n+3}{n^2-7n+10}\right)^{n+\sin(n!)}$$

Unfortunately $n^2-4n+3$ (the numerator) and $n^2-7n+10$ (the denominator) don't have a common root (or common roots). So here's my try $$\lim_{n\to\infty}\left(\dfrac{n^2-4n+3}{n^2-7n+10}\right)^{n+\sin(n!)}\\=\lim_{n\to\infty}\left(\dfrac{n^2-7n+10+3n-7}{n^2-7n+10}\right)^{n+\sin(n!)}\\=\lim_{n\to\infty}\left(1+\dfrac{3n-7}{n^2-7n+10}\right)^n\cdot\left(1+\dfrac{3n-7}{n^2-7n+10}\right)^{\sin(n!)}\\=\lim_{n\to\infty}\left(1+\dfrac{\frac{3n^2-7n}{n^2-7n-10}}{n}\right)^n\left(1+\dfrac{3n-7}{n^2-7n+10}\right)^{\sin(n!)}\\=e^3\lim_{n\to\infty}\left(1+\dfrac{n\left(3-\frac{7}{n}\right)}{n^2\left(1-\frac{7}{n}+\frac{10}{n^2}\right)}\right)^{\sin(n!)}$$

Can we say that this limit is $1$ as $\dfrac{n\left(3-\frac{7}{n}\right)}{n^2\left(1-\frac{7}{n}+\frac{10}{n^2}\right)}\to0$ as $n\to\infty$ or this is not so clear (and might not be true in all cases)? ( So the initial limit maybe is $e^3\cdot1=e^3$. )

Answer 1

As you pointed out, it suffices to show $$\lim_{n\to\infty}\left(1+\dfrac{3n-7}{n^2-7n+10}\right)^{\sin(n!)} = 1$$ to conclude your limit is $e^3$. Remark that $\frac{3n - 7}{n^2 -7n + 10} > 0$ for $n > n_0$ large enough (for $n$ large enough, both the numerator and denominator are $>0$). So, for $n > n_0$, we have $$\left(1+\dfrac{3n-7}{n^2-7n+10}\right)^{-1} \leq \left(1+\dfrac{3n-7}{n^2-7n+10}\right)^{\sin(n!)} \leq \left(1+\dfrac{3n-7}{n^2-7n+10}\right)^{1}$$ because $-1 \leq \sin(n!) \leq 1$. The limit then follows from the sandwich theorem because the outer sides of the inequality both converge to 1.

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I used the fact that if $\alpha > 1$, then $x \mapsto \alpha^x$ is an increasing function.

Answer 2

• Let us take $\ln$ of this sequence, because it is easier to get first an asymtotic equivalent of a product.

  We shall use that $n+\sin(n!)\sim n$ (since $|\sin(n!)|\le1$), and that $\ln(1+u)\sim u$ as $u\to0.$ $$\begin{align}\ln\left(\left(\dfrac{n^2-4n+3}{n^2-7n+10}\right)^{n+\sin(n!)}\right)&=\left(n+\sin(n!)\right)\left(\ln\left(1-\frac4n+\frac3{n^2}\right)-\ln\left(1-\frac7n+\frac{10}{n^2}\right)\right)\\&\sim n\left(-\frac4n+\frac7n\right)\\&\to3,\end{align}$$ hence $$\left(\dfrac{n^2-4n+3}{n^2-7n+10}\right)^{n+\sin(n!)}\to e^3.$$

• Your guess that $\lim_{n\to\infty}\left(1+\dfrac{n\left(3-\frac{7}{n}\right)}{n^2\left(1-\frac{7}{n}+\frac{10}{n^2}\right)}\right)^{\sin(n!)}=1$ was correct, and your argument $\dfrac{n\left(3-\frac{7}{n}\right)}{n^2\left(1-\frac{7}{n}+\frac{10}{n^2}\right)}\to0$ was nearly sufficient. More generally, if $a_n\to1$ and $(b_n)$ is bounded then $a_n^{b_n}\to1,$ since $\ln\left(a_n^{b_n}\right)=b_n\ln(a_n)\to0$ by the squeeze theorem.

• Your claim that $\lim_{n\to\infty}\left(1+\dfrac{\frac{3n^2-7n}{n^2-7n-10}}{n}\right)^n=e^3$ is also correct, but does not follow from $\lim_{n\to\infty}\left(1+\dfrac3n\right)^n=e^3$. It needs some analogous argument: if $c_n\to c$ then $\left(1+\dfrac{c_n}n\right)^n\to e^c,$ since $\ln\left(\left(1+\dfrac{c_n}n\right)^n\right)=n\ln\left(1+\dfrac{c_n}n\right)\sim n\frac{c_n}n=c_n\to c.$