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Limit of the function
:
Limit of sin(3)^2/x
Limit of x^4+2*cos(x)^2
Limit of x*2^x*3^(-x)
Limit of log(1+3*x^2)/(x^3-5*x^2)
Derivative of
:
-x^2
Graphing y =
:
-x^2
Integral of d{x}
:
-x^2
Identical expressions
-x^ two
minus x squared
minus x to the power of two
-x2
-x²
-x to the power of 2
Similar expressions
(4-x^2)/(3-x^2)
1/(9-x^2)
(1-x^2)/(1+x^2)
x^2
1-6*x-x^2/3
log(1-x^2)/x^2
Limit of the function
/
-x^2
Limit of the function -x^2
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]
/ 2\ lim \-x / x->oo
$$\lim_{x \to \infty}\left(- x^{2}\right)$$
Limit(-x^2, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(- x^{2}\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to \infty}\left(- x^{2}\right)$$ =
$$\lim_{x \to \infty} \frac{1}{\left(-1\right) \frac{1}{x^{2}}}$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty} \frac{1}{\left(-1\right) \frac{1}{x^{2}}} = \lim_{u \to 0^+}\left(- \frac{1}{u^{2}}\right)$$
=
$$- \frac{1}{0} = -\infty$$
The final answer:
$$\lim_{x \to \infty}\left(- x^{2}\right) = -\infty$$
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
$$\lim_{x \to \infty}\left(- x^{2}\right) = -\infty$$
$$\lim_{x \to 0^-}\left(- x^{2}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(- x^{2}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(- x^{2}\right) = -1$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(- x^{2}\right) = -1$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(- x^{2}\right) = -\infty$$
More at x→-oo
The graph