# 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?

• This is a standard problem Take the log, pull out the exponent, invert the exponent to the denominator, L'hospital's, exponentiate. Done. – B. Goddard Jan 3 '17 at 0:07
• Hint: $\lim_{n\to\infty}\left(\frac{-5+3n}{3n}\right)^{6n} = \lim_{n\to\infty}\left(1 - \frac{5}{3n}\right)^{6n}$. – user361424 Jan 3 '17 at 0:08
• This would be a ridiculous problem to use l'Hôpital's. – user361424 Jan 3 '17 at 0:09
• Duplicate: $\displaystyle\lim_{m\to\infty}\left(1+\frac{r}{m}\right)^{mt}$ ($r=-\tfrac53,t=6$). (Found using Approach0.xyz) – Workaholic Jan 3 '17 at 15:08

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 $x$.

$\left\{\left(1-\frac{5}{3n}\right)^{6n}\right\}_{n\geq 2}$ 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.

Hint: write it as follows and note that the term inside the big parenthesis has limit $e\,$:

$$\lim_{n\to\infty} \left( \left(1 + \frac{1}{\frac{-3n}{5}}\right)^{\frac{-3n}{5}} \right)^{-10}$$

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

$$\lim_{n\rightarrow \infty} \left(\frac{-5+3n}{3n}\right)^{6n} = \left(\lim_{n\rightarrow \infty}\left(1+\frac{\frac{-5}{3}}{n}\right)^n\right)^6 = \left(e^{\frac{-5}{3}}\right)^6 = e^{-10}$$

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.

$\lim\limits_{n\to\infty}(\frac{-5+3n}{3n})^{6n}$=$\lim\limits_{n\to\infty}(1+\frac{-5}{3n})^{6n}$=$\lim\limits_{n\to\infty}((1+\frac{1}{\frac{3n}{-5}})^\frac{3n}{-5})^{-10}$

let t=$\frac{3n}{-5}$,we have:

$\lim\limits_{n\to\infty}(\frac{-5+3n}{3n})^{6n}$=$\lim\limits_{t\to\infty}((1+\frac{1}{t})^t)^{-10}$=$e^{-10}$