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Limit of the function
:
Limit of 7^(1/(-3+x))
Limit of (3-3*x^2+4*x^4+6*x^3)/(2*x^2+7*x^4)
Limit of ((5+4*x)/(-1+5*x))^(1+3*x)
Limit of (-6-x^2-3*x+4*x^3)/(3-x^2+2*x^3)
Derivative of
:
5/x^3
Graphing y =
:
5/x^3
Identical expressions
five /x^ three
5 divide by x cubed
five divide by x to the power of three
5/x3
5/x³
5/x to the power of 3
5 divide by x^3
Limit of the function
/
5/x^3
Limit of the function 5/x^3
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]
/5 \ lim |--| x->oo| 3| \x /
$$\lim_{x \to \infty}\left(\frac{5}{x^{3}}\right)$$
Limit(5/x^3, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(\frac{5}{x^{3}}\right)$$
Let's divide numerator and denominator by x^3:
$$\lim_{x \to \infty}\left(\frac{5}{x^{3}}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{5 \frac{1}{x^{3}}}{1}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{5 \frac{1}{x^{3}}}{1}\right) = \lim_{u \to 0^+}\left(5 u^{3}\right)$$
=
$$5 \cdot 0^{3} = 0$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{5}{x^{3}}\right) = 0$$
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
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to \infty}\left(\frac{5}{x^{3}}\right) = 0$$
$$\lim_{x \to 0^-}\left(\frac{5}{x^{3}}\right) = -\infty$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{5}{x^{3}}\right) = \infty$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{5}{x^{3}}\right) = 5$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{5}{x^{3}}\right) = 5$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{5}{x^{3}}\right) = 0$$
More at x→-oo
Rapid solution
[src]
0
$$0$$
Expand and simplify
The graph