<s>[INST] 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.
 [/INST] Take $$\ln f(x)=({x^3+3x})\ln\left(\frac{x^4+\sqrt{x^2 +1}}{x^4-x+2}\right)$$
and note
$$\frac{x^4+\sqrt{x^2 +1}}{x^4-x+2}
=\frac{1+\frac1{x^3}\sqrt{1+\frac1{x^2}}}{1-\frac1{x^3}+\frac2{x^4}}
=\frac{1+\frac1{x^3}+O(\frac1{x^5})}{1-\frac1{x^3}+\frac2{x^4}}
=1+ \frac2{x^3}+O(\frac1{x^4})$$
Then,
$$\lim_{x\to\infty} \ln f(x)=\lim_{x\to\infty}(x^3+3x)\ln(1+ \frac2{x^3}+O(\frac1{x^4}))$$
$$=\lim_{x\to\infty} (x^3+3x)(\frac2{x^3}+O(\frac1{x^4}))=\lim_{x\to\infty} (2+O(\frac1{x}))=2$$
Thus, 
$$\lim_{x\to\infty} f(x)=e^2$$
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