Determine $\lim_{x\to\infty} f(x)$, where $f(x)=\left(\frac{x^4+(x^2 +1)^{0.5}}{x^4-x+2}\right)^{x^3+3x}$ 
Determine $\lim_{x\to\infty} f(x)$ where
  $$f(x)=\left(\frac{x^4+(x^2 +1)^{0.5}}{x^4-x+2}\right)^{x^3+3x}.$$

I used the binomial expansion method and my answer is $e^2$ but when I plot this function on Desmos and check value of this function at very large values like $10^9$ it shows function is equal to $1$.
I am more concerned about why this happens rather than answer.

We have that
$$\left(\frac{x^4+(x^2 +1)^{0.5}}{x^4-x+2}\right)^{x^3+3x}=\left[\left(1+\frac{(x^2 +1)^{0.5}+x-2}{x^4-x+2}\right)^{\frac{x^4-x+2}{(x^2 +1)^{0.5}+x-2}}\right]^{(x^3+3x)\frac{(x^2 +1)^{0.5}+x-2}{x^4-x+2}}$$
and
$$\left(1+\frac{(x^2 +1)^{0.5}+x-2}{x^4-x+2}\right)^{\frac{x^4-x+2}{(x^2 +1)^{0.5}+x-2}} \to e$$
then we need to evaluate
$$\lim_{x\to \infty} \frac{(x^3+3x)[(x^2 +1)^{0.5}+x-2]}{x^4-x+2}$$
which turns to be equal to $2$, therefore your result seems correct.
