How to solve this lim?

Question

$\lim _{n \rightarrow+\infty}(n-3)(3 \ln (n-1)-\ln (n+2)-2 \ln (n+1))$

I tried to get all in one ln but then I get this: $\ln \left(\lim _{n\rightarrow \infty}\left(\frac{(n-1)^{3}}{(n+2)(n+1)^{2}}\right)^{(n-3)}\right)$

$\ln \left(\lim _{n \rightarrow \infty}\left(\frac{\left(n^{3}-3 n^{2}+3 n-1\right)^{(n-3)}}{\left(n^{3}+4 n^{2}+5 n+2\right)^{(n-3)}}\right)\right)$

Have no idea how to get out of this..

Answer 1

The limit is $-7$. Take $x=1/n$ with $x\to 0^+$. After making the substitution and simplifying you get \begin{eqnarray*} && \lim_{n\to+\infty}(n-3)(3\ln (n-1)-\ln (n+2)-2\ln (n+1))\\ & = & \lim _{x\to 0^+}\left(\frac{1}{x}-3\right)(3\ln(1-x)-\ln(1+2x)-2\ln(1+x))\\ & = & \lim _{x\to 0^+}(1-3x)\left(-3\frac{\ln(1-x)}{-x}-2\frac{\ln(1+2x)}{2x}-2\frac{\ln(1+x)}{x}\right) \\ & = & -7\end{eqnarray*} since $\lim_{t\to 0}\frac{\ln(1+t)}{t}=1$

Answer 2

It is $1^\infty$ form. (Cubic/cubic with same coefficient of $x^3$ ). There is a direct formula for it. $\lim_{x\rightarrow \infty} f(x)^{g(x)} = e^{\lim_{x \rightarrow \infty } (f(x)-1)g(x)}$ where $\lim_{x\rightarrow \infty} f(x) =1$ and $\lim_{x\rightarrow \infty} g(x) =\infty$

Answer 3

$$\lim_{n\rightarrow\infty}(n-3)(3\log(n-1)-2\log(n+1)-\log(n+2))$$ $$=\lim_{n\rightarrow\infty}(n-3)(3\log(n-1)-3\log(n)+3\log(n)-2\log(n+1)-\log(n+2))$$ $$=\lim_{n\rightarrow\infty}3\log((1-\frac{1}{n})^{n-3})-2\log((1+\frac{1}{n})^{n-3})-\log((1+\frac{2}{n})^{n-3})$$ $$=3\log(\lim_{n\rightarrow\infty}(1-\frac{1}{n})^{n-3})-2\log(\lim_{n\rightarrow\infty}(1+\frac{1}{n})^{n-3})-\log(\lim_{n\rightarrow\infty}(1+\frac{2}{n})^{n-3})$$$$=3\log(e^{-1})-2\log(e)-\log(e^{2})=-3-2-2=-7$$

Answer 4

The expression under limit can be written as $$(n-3)\log\frac{(n-1)^3}{(n+2)(n+1)^2}=(n-3)\log\frac{n^3-3n^2+\cdots}{n^3+4n^2+\cdots} $$ and this gets expressed further as $$ (n-3)\log\left(1-\frac{7n^2+\dots}{n^3+4n^2+\dots}\right)$$ and using the limit $\lim\limits _{x\to 0} \dfrac {\log(1+x)}{x}=1$ the limit of the expression in question is same as that of expression $$-(n-3)\cdot\frac{7n^2+\dots}{n^3+4n^2+\dots}$$ which is $-7$.