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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?

  • 2
    This is a standard problem Take the log, pull out the exponent, invert the exponent to the denominator, L'hospital's, exponentiate. Done. – 2017-01-03
  • 0
    Hint: $\lim_{n\to\infty}\left(\frac{-5+3n}{3n}\right)^{6n} = \lim_{n\to\infty}\left(1 - \frac{5}{3n}\right)^{6n}$. – 2017-01-03
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    This would be a ridiculous problem to use l'Hôpital's. – 2017-01-03
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    Duplicate: [$\displaystyle\lim_{m\to\infty}\left(1+\frac{r}{m}\right)^{mt}$](https://math.stackexchange.com/q/244540/201168) ($r=-\tfrac53,t=6$). (*Found using [Approach0.xyz](https://approach0.xyz/search/?q=%24%5Clim_%7Bn%5Cto%5Cinfty%7D(1%2B%5Cfrac%7Ba%7D%7Bn%7D)%5E%7Bbn%7D%24&p=1)*) – 2017-01-03

6 Answers 6

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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 $x$.

3

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

2

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

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

2

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

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

0

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

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

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

0

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.