<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] 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.)  
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