Find $\lim\limits_{n\to\infty}\left(\frac{-5+3n}{3n}\right)^{6n}$ I want to find the limit of the following:
$$\lim_{n\to\infty}\left(\frac{-5+3n}{3n}\right)^{6n}$$
However, although I see the quite similar structure to $\left(1+\frac{1}{n}\right)^n$, I do not come to a meaningful result. What is the limit?
 A: Hint
Note that 
$$
\left(\frac{-5+3n}{3n}\right)^{6n}=\left(\left(1+\frac{-5/3}{n}\right)^{n}\right)^6
$$
and use the fact that
$$
\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x
$$
for all $x$.
A: $\left\{\left(1-\frac{5}{3n}\right)^{6n}\right\}_{n\geq 2}$ is a subsequence of $\left\{\left(1-\frac{5}{n}\right)^{2n}\right\}_{n\geq 6}$ .
Since $\left(1-\frac{5}{n}\right)^n$ converges to $e^{-5}$ as $n\to +\infty$, $\left(1-\frac{5}{n}\right)^{2n}$ converges to $e^{-10}$.
Finish with: the subsequences of a converging sequence share the same limit.
A: Hint: write it as follows and note that the term inside the big parenthesis has limit $e\,$:
$$\lim_{n\to\infty} \left( \left(1 + \frac{1}{\frac{-3n}{5}}\right)^{\frac{-3n}{5}} \right)^{-10}$$
A: The limit you recalled evaluated to $e$. Which similar one evaluates to $e^x$? That would be my hint.
Complete solution
$$\left(\frac{-5+3n}{3n}\right)^{6n}= \left(\left(1+\frac{\frac{-5}{3}}{n}\right)^n\right)^6$$
$x^a$ is continuous for any positive $a$, hence we can switch it with $\lim$. Therefore
$$\lim_{n\rightarrow \infty} \left(\frac{-5+3n}{3n}\right)^{6n} = \left(\lim_{n\rightarrow \infty}\left(1+\frac{\frac{-5}{3}}{n}\right)^n\right)^6 = \left(e^{\frac{-5}{3}}\right)^6 = e^{-10}$$
A: $\lim\limits_{n\to\infty}(\frac{-5+3n}{3n})^{6n}$=$\lim\limits_{n\to\infty}(1+\frac{-5}{3n})^{6n}$=$\lim\limits_{n\to\infty}((1+\frac{1}{\frac{3n}{-5}})^\frac{3n}{-5})^{-10}$
let  t=$\frac{3n}{-5}$,we have:
$\lim\limits_{n\to\infty}(\frac{-5+3n}{3n})^{6n}$=$\lim\limits_{t\to\infty}((1+\frac{1}{t})^t)^{-10}$=$e^{-10}$
A: Another possible solution.
Consider $$A=\left(\frac{-5+3n}{3n}\right)^{6n}\implies \log(A)=6n\log\left(1-\frac{5}{3n}\right)$$ Since $n$ is large, use Taylor series $$\log(1-x)=-x-\frac{x^2}{2}+O\left(x^3\right)$$ Replace $x=\frac{5}{3n}$ to get $$\log\left(1-\frac{5}{3n}\right)=-\frac{5}{3 n}-\frac{25}{18 n^2}+O\left(\frac{1}{n^3}\right))$$ $$\log(A)=-10-\frac{25}{3 n}+O\left(\frac{1}{n^2}\right)$$ Now, use Taylor again $$A=e^{\log(A)}=\frac{1}{e^{10}}-\frac{25}{3 e^{10} n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.
