Computing the limit of $ \frac{1^3+2^3+\cdots+n^3}{\sqrt{4n^8 +1}} $ I had this exercise:

Compute the limit
$$ \lim_{n\to\infty} \frac{1^3+2^3+\cdots+n^3}{\sqrt{4n^8 +1}} $$

I tried two different approaches and got different answers.
Approach 1:
$$\begin{split}
\lim_{n\to\infty} \frac{1^3+2^3+\cdots+n^3}{n^4\sqrt{4 +1/n^8}}&=\lim_{n\to\infty} \frac{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\cdots+\frac{n^3}{n^4}}{\sqrt{4 +1/n^8}}\\
&=\lim_{n\to\infty} \frac{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\cdots+\frac{1}{n}}{\sqrt{4 +1/n^8}}\\
&= \frac{0}{\sqrt{4+0}}\\
&= 0\\
\end{split}$$
Approach 2:
We substitute sum of cubes of $n$ natural numbers in the numerator and get
$$\begin{split}
\lim_{n\to\infty} \frac{n^2(n+1)^2}{4n^4\sqrt{4 +1/n^8}}&= \lim_{n\to\infty} \frac{n^2(n+1)^2}{4n^4\sqrt{4 +1/n^8}}\\
&= \lim_{n\to\infty} \frac{n^4(1+\frac{1}{n})^2}{4n^4\sqrt{4 +1/n^8}}\\
&= \lim_{n\to\infty} \frac{(1+\frac{1}{n})^2}{4\sqrt{4 +1/n^8}}\\ 
&= \lim_{n\to\infty} \frac{1}{4\sqrt{4}}\\
&= \frac{1}{8}\\
\end{split}$$
I'm not sure why the two methods are giving me different answers and which one is correct?

One more method is to see the Laurent series of the expression. It is equal to $$\frac{1}{8}+\frac{1}{4n}+\frac{1}{8n^2}-\frac{1}{64n^8}+O\left(\left(\frac{1}{n}\right)^9\right)$$
You can see that it confirms your second approach.
