Evaluating the limit $\lim_{x\rightarrow\infty}\frac{x^4(1-x^4)}{1+x^2}$

Question

I want to evaluate the following limit:

$$\lim_{x\rightarrow\infty}\frac{x^4(1-x^4)}{1+x^2}$$

Thanks for any help.

Answer 1

Solving:

$$\lim_{x\to\infty}\frac{x^4(1-x^4)}{1+x^2}$$

Since $x^4(1-x^4)$ grows asymptotically faster than $1+x^2$ as $x$ approaches $\infty$; $\lim_{x\to\infty}\frac{x^4(1-x^4)}{1+x^2}\to\pm\infty$. Additionally $x^4(1-x^4)<0$ and $1+x^2>0$ as $x$ approaches $\infty$ so:

$$\lim_{x\to\infty}\frac{x^4(1-x^4)}{1+x^2}\to-\infty$$

Answer 2

Simplify your expression to $x^4(1-x^2)$. Then the limit is $-\infty$.

Answer 3

$\lim_{x\rightarrow\infty}\frac{x^4(1-x^4)}{1+x^2}$

$\implies \frac{x^{4}(1-x^{2})(1+x^{2})}{(1+x^{2})}$

$\implies x^{4}(1-x^{2})$

Take $x^{2} out : $

$\implies x^{4}.x^{2}(\frac{1}{x^{2}}-1)$

Now apply the limit as $ lim_{x\to\infty}$

$\implies x ^{6} (-1) \implies -\infty$