Trouble evaluating the $\lim_{n\rightarrow \infty}(\frac{n-1}{n+2})^{n+2}$ I am reviewing my calculus and am not sure why I got this limit incorrectly, I know via wolfram it should be $e^{-3}$.
$$L=\lim_{n\rightarrow \infty}(\frac{n-1}{n+2})^{n+2}\Rightarrow \log(L)=\lim_{n\rightarrow \infty}(n+2)[\log(n-1)-\log(n+2)]\sim \lim_{n\rightarrow \infty}(n+2)[\log(n)-\log(n)]$$ via first order taylor approx around $n$
Which should give a limit of $\log(L)=0$ and $L=1$
What assumption should I have not made? How do you evaluate this limit? 
 A: Here's a simple and quick method:
Use $$\lim_{n\rightarrow \infty}\left(1+\frac1n\right)^n=e$$
Then
$$\lim_{n\rightarrow \infty}(\frac{n-1}{n+2})^{n+2}=\lim_{n\rightarrow \infty}(\frac{n+2-3}{n+2})^{n+2}=\lim_{n\rightarrow \infty}(1-\frac{3}{n+2})^{n+2}=e^{-3}=\frac1{e^3}$$
A: One may recall the following standard limit, for any $x \in \mathbb{R}$, as $N \to \infty$,
$$
\left(1+\frac{x}N \right)^N \to e^x. \tag1
$$ Then, by writting
$$
\left(\frac{n-1}{n+2}\right)^{n+2}=\left(\frac{(n+2)-3}{n+2}\right)^{n+2}=\left(1+\frac{x}N \right)^N
$$ one may apply $(1)$ with $x=-3$ and $ N=n+2$ obtaining $e^{-3}$ as the sought limit.
A: Hint:
$$L=\lim_{n\rightarrow \infty}(\frac{n-1}{n+2})^{n+2}\Rightarrow \log(L)=\lim_{n\rightarrow \infty}(n+2)[\log(n-1)-\log(n+2)] = \lim_{n\rightarrow \infty}(n+2)[(\log(n) + \log(1-1/n)) - (\log(n)+\log(1-2/n)] $$ 
Taylor expand $\log$ ...
A: You approximated too bluntly in replacing $\log(n-1)-\log(n+2)$ by $\log(n)-\log(n)$. The problem here is that though these are separately close to $\log(n)$, they very nearly cancel each out, and only higher order terms survive. Instead, you can use Taylor approximation to get:
$$\log(n-1)-\log(n+2)=\log(n)-1/n-(\log(n)+2/n))+o(1/n)=-3/n+o(1/n).$$
Then you multiply by the $n+2$ on the outside and continue.
Provided you could justify saying that $\log(n-1)-\log(n+2) \to 0$, you could also use L'Hopital's rule, by writing
$$(n+2)(\log(n-1)-\log(n+2))=\frac{\log(n-1)-\log(n+2)}{\frac{1}{n+2}}.$$
A: Using only "elementary arguments" (continuity of $\exp$, and a well-known limit).
You can rewrite the quantity you are interested in in the (often more convenient) exponential form:
$$
\left(\frac{n-1}{n+2}\right)^{n+2} 
= e^{(n+2)\ln\left(\frac{n-1}{n+2}\right)}
= e^{(n+2)\ln\left(1-\frac{3}{n+2}\right)}
= e^{-3\cdot\left(-\frac{n+2}{3}\ln\left(1-\frac{3}{n+2}\right)\right) }
$$
Now, using the fact that
$$
\frac{\ln(1+x)}{x} \xrightarrow[x\to 0]{} 1
$$
(proven e.g. by recognizing the derivative of $f(x) = \ln(1+x)$ at $0$), you have that 
$$
-\frac{n+2}{3}\ln\left(1-\frac{3}{n+2}\right) \xrightarrow[n\to\infty]{} 1
$$
since $-\frac{3}{n+2}\xrightarrow[n\to\infty]{} 0$. By continuity of exponential, you thus get
$$
e^{-3\cdot\left(-\frac{n+2}{3}\ln\left(1-\frac{3}{n+2}\right)\right) }
\xrightarrow[n\to\infty]{} e^{-3\cdot 1} = e^{-3}.
$$
A: The step in which you made an approximation was not correct. The usual way to prove this uses continuity and $\exp \circ \log$, as follows.
$$\lim_{n\to\infty} {\left(\frac{n-1}{n+2}\right)}^{n+2}=\lim_{n\to\infty} {\left(1-\frac{3}{n+2}\right)}^{n+2}$$
By continuity, that limit is equal to:
$$\exp\left(\lim_{n\to\infty} (n+2)\cdot \ln\left(1-\frac{3}{n+2}\right)\right)$$
If you know the power series expansion of $\ln(1+x)$ the answer should be immediate from the $-3$ in the numerator, but if you don't we can still solve this using L'Hopital. Rewrite the limit inside the exponential as:
$$\lim_{n\to\infty} \frac{\ln\left(1-\frac{3}{n+2}\right)}{\frac{1}{n+2}}$$
Which is now an indeterminate form of type $\frac{0}{0}$. Applying L'Hopital, that limit is equal to
$$\lim_{n\to\infty} \frac{\displaystyle\frac{3}{{(n+2)}^2\left(1-\frac{3}{n+2}\right)}}{\displaystyle-\frac{1}{{(n+2)}^2}}=\lim_{n\to\infty}\frac{-3}{\left(1-\frac{3}{n+2}\right)}=-3$$
So that the original limit is $\exp(-3)$, as desired.
