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Limit of the function
:
Limit of (-1+e^(2*x))/(3*x)
Limit of (-1+3*x)/(5+x^2+7*x)
Limit of 1+13*x/5
Limit of (7+x+x^2)/(-1+e^x)
Derivative of
:
4*x^3
Integral of d{x}
:
4*x^3
Identical expressions
four *x^ three
4 multiply by x cubed
four multiply by x to the power of three
4*x3
4*x³
4*x to the power of 3
4x^3
4x3
Similar expressions
(1+2*x+2*x^2)/(2-x+4*x^3)
(-5*x+4*x^3)/(1-3*x^3)
(1-6*x^3)^(1/(4*x^3))
cot(4*x)^3*sin(3*x)^3
(x+4*x^3)/(3-x+2*x^2)
Limit of the function
/
4*x^3
Limit of the function 4*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]
/ 3\ lim \4*x / x->-oo
$$\lim_{x \to -\infty}\left(4 x^{3}\right)$$
Limit(4*x^3, x, -oo)
Detail solution
Let's take the limit
$$\lim_{x \to -\infty}\left(4 x^{3}\right)$$
Let's divide numerator and denominator by x^3:
$$\lim_{x \to -\infty}\left(4 x^{3}\right)$$ =
$$\lim_{x \to -\infty} \frac{1}{\frac{1}{4} \frac{1}{x^{3}}}$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to -\infty} \frac{1}{\frac{1}{4} \frac{1}{x^{3}}} = \lim_{u \to 0^+}\left(\frac{4}{u^{3}}\right)$$
=
$$\frac{4}{0} = -\infty$$
The final answer:
$$\lim_{x \to -\infty}\left(4 x^{3}\right) = -\infty$$
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
$$\lim_{x \to -\infty}\left(4 x^{3}\right) = -\infty$$
$$\lim_{x \to \infty}\left(4 x^{3}\right) = \infty$$
More at x→oo
$$\lim_{x \to 0^-}\left(4 x^{3}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(4 x^{3}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(4 x^{3}\right) = 4$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(4 x^{3}\right) = 4$$
More at x→1 from the right
The graph