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Limit of the function
:
Limit of (7+n)/(5+n)
Limit of (9+3*x^2+4*x)/(7-7*x+3*x^2)
Limit of (1+3*x^2)/(2+x^2)
Limit of ((3-n)^4-(2-n)^4)/((1-n)^3-(1+n)^3)
Derivative of
:
-x^4
Integral of d{x}
:
-x^4
Identical expressions
-x^ four
minus x to the power of 4
minus x to the power of four
-x4
-x⁴
Similar expressions
(3+x^5-x^4)/(1+x+x^2+x^5)
(x^5-x^4)/(1-x)
2-x^2-5*x^3+2*x^5-x^4/3
(x^8-x^7)/(x^5-x^4)
1-x-x^4+5*x^2
x^4
Limit of the function
/
-x^4
Limit of the function -x^4
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->oo
$$\lim_{x \to \infty}\left(- x^{4}\right)$$
Limit(-x^4, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(- x^{4}\right)$$
Let's divide numerator and denominator by x^4:
$$\lim_{x \to \infty}\left(- x^{4}\right)$$ =
$$\lim_{x \to \infty} \frac{1}{\left(-1\right) \frac{1}{x^{4}}}$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty} \frac{1}{\left(-1\right) \frac{1}{x^{4}}} = \lim_{u \to 0^+}\left(- \frac{1}{u^{4}}\right)$$
=
$$- \frac{1}{0} = -\infty$$
The final answer:
$$\lim_{x \to \infty}\left(- x^{4}\right) = -\infty$$
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
$$\lim_{x \to \infty}\left(- x^{4}\right) = -\infty$$
$$\lim_{x \to 0^-}\left(- x^{4}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(- x^{4}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(- x^{4}\right) = -1$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(- x^{4}\right) = -1$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(- x^{4}\right) = -\infty$$
More at x→-oo
The graph