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