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


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

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

A: 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.)  
