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x^4-x

Limit of the function x^4-x

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     / 4    \
 lim \x  - x/
x->oo        
$$\lim_{x \to \infty}\left(x^{4} - x\right)$$
Limit(x^4 - x, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(x^{4} - x\right)$$
Let's divide numerator and denominator by x^4:
$$\lim_{x \to \infty}\left(x^{4} - x\right)$$ =
$$\lim_{x \to \infty}\left(\frac{1 - \frac{1}{x^{3}}}{\frac{1}{x^{4}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{1 - \frac{1}{x^{3}}}{\frac{1}{x^{4}}}\right) = \lim_{u \to 0^+}\left(\frac{1 - u^{3}}{u^{4}}\right)$$
=
$$\frac{1 - 0^{3}}{0} = \infty$$

The final answer:
$$\lim_{x \to \infty}\left(x^{4} - x\right) = \infty$$
Lopital's rule
There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type
The graph
Rapid solution [src]
oo
$$\infty$$
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to \infty}\left(x^{4} - x\right) = \infty$$
$$\lim_{x \to 0^-}\left(x^{4} - x\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(x^{4} - x\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(x^{4} - x\right) = 0$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(x^{4} - x\right) = 0$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(x^{4} - x\right) = \infty$$
More at x→-oo
The graph
Limit of the function x^4-x