<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 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

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

</s>