<s>[INST] 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.
 [/INST] 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 $e^{-5*2018^4}$
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