How to find $\lim_\limits{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}$

Question

How to find $$\lim_{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}$$

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My attempt:

$$\lim _{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}=\lim _{n\to \infty }\left(1-\frac{5}{n^4}\right)^{\left(4\cdot 2018n+4\right)}=\lim _{n\to \infty }\left(\left(1-\frac{5}{n^4}\right)^{\left(4\cdot 2018n\right)}+\left(1-\frac{5}{n^4}\right)^4\right)=\lim _{n\to \infty }\left(\left(\left(1-\frac{5}{n^4}\right)^{n}\right)^{4*2018}+\left(1-\frac{5}{n^4}\right)^4\right)$$

so we get :

• $\lim _{n\to \infty }\left(1-\frac{5}{n^4}\right)^4=1$

let:

$b_n=\frac{-5}{n^3}$

$\lim _{n\to \infty }\frac{-5}{n^3}=0$

• $\lim _{n\to \infty }\left(\left(1-\frac{5}{n^4}\right)^n\right)^{4\cdot 2018}=\lim \:_{n\to \:\infty \:}\left(\left(1+\frac{b_n}{n}\right)^n\right)^{4\cdot \:2018}=\left(e^0\right)^{4\cdot 2018}=1$

so the answer is :

$$\lim _{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}=1+1=2$$

Is this answer correct if not how can I find the limit ?

thanks.

Answer 1

The standard way is to consider

$$(1-\frac{5}{n^4})^{(2018n+1)^4}=e^{(2018n+1)^4 \log{(1-\frac{5}{n^4}) }}\to e^{-5\cdot2018^4}$$

indeed by standard limits

$${(2018n+1)^4 \log{(1-\frac{5}{n^4}) }}=(2018n+1)^4(\frac{5}{n^4}) \frac{\log {(1-\frac{5}{n^4})}}{\frac{5}{n^4}}\to -5\cdot2018^4$$

Or as an alternative

$$(1-\frac{5}{n^4})^{(2018n+1)^4}=\left[(1-\frac{5}{n^4})^{\frac{n^4}5}\right]^{\frac5{n^4}(2018n+1)^4}\to e^{-5\cdot2018^4}$$

Answer 2

The result seems to me weird. You wrote $\displaystyle \left(2018n+1\right)^4=4\left(2018n+1\right)$ which is false. And you when an expression is at power $n$, you cannot say the argument tends to $0$ hence it tends to $0^n$.

I would rather write $$ \left(1-\frac{5}{n^4}\right)^{\left(2018n+1\right)^4}=e^{\left(2018n+1\right)^4\ln\left(1-\frac{5}{n^4}\right)} $$ Then $$ \ln\left(1-\frac{5}{n^4}\right)\underset{(+\infty)}{=}-\frac{5}{n^4}+o\left(\frac{1}{n^4}\right) $$ and $$ \left(2018n+1\right)^4=2018^4n^4+o\left(n^4\right)=16583822760976n^4+o\left(n^4\right)$$ Finally

$$ \left(1-\frac{5}{n^4}\right)^{\left(2018n+1\right)^4}\underset{n \rightarrow +\infty}{\rightarrow}e^{-82919113804880}$$

Answer 3

You can also write

$$(1-5/n^4)^{(2018n +1)^4} = [(1-5/n^4)^{n^4}]^{(2018n +1)^4/n^4}.$$

Inside the brackets we have the limit $e^{-5},$ while the outer exponent $\to 2018^4.$ Thus the desired limit is $e^{-5\cdot 2018^4}.$ (We have used the result: If $a_n \to a\in (0,\infty)$ and $b_n\to b\in \mathbb R,$ then $a_n^{b_n} \to a^b.$ This is worthwhile proving for yourself, as it comes up often.)

Answer 4

Let $m=2018n+1$ then $\lim _{m\to \infty }\big(1-5\frac{2018^4}{(m-1)^4}\big)^{m^4}$

=$\lim _{m\to \infty }\big(1-5\frac{2018^4}{(m-1)^4}\big)^{{m^4}\frac{(m-1)^4}{(m-1)^4}}$

=$\lim _{m\to \infty }\big(1-5\frac{2018^4}{(m-1)^4}\big)^{{(m-1)^4}{\frac{1}{(1-\frac{1}{m})^4}}}$

Known that $\lim _{m\to \infty }\big(1-\frac{x}{m}\big)^{m}=e^{-x}$

The limit is equal to $e^{-5*2018^4}$