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Limit of the function
:
Limit of (-1+e^x)/sin(x)
Limit of (5+x^2-6*x)/(-1+x^2)
Limit of ((-1+x)/(4+x))^(2+3*x)
Limit of (1+n)/(2+n)
Factor polynomial
:
x^4-x
Integral of d{x}
:
x^4-x
Graphing y =
:
x^4-x
Identical expressions
x^ four -x
x to the power of 4 minus x
x to the power of four minus x
x4-x
x⁴-x
Similar expressions
x^4+x
1+x^4-x^3-4*x+2/x^3
(x^4-x^2)/(x^2*log(10))
(x^5-x)/(1+x^4-x^2+5*x^3)
(5*x^2+7*x)/(7+x^4-x^2)
Limit of the function
/
x^4-x
Limit of the function x^4-x
at
→
Calculate the limit!
v
For end points:
---------
From the left (x0-)
From the right (x0+)
The graph:
from
to
Piecewise:
{
enter the piecewise function here
The solution
You have entered
[src]
/ 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
Plot the graph
Rapid solution
[src]
oo
$$\infty$$
Expand and simplify
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