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Limit of the function
:
Limit of ((-2+x)/(3+x))^(4-x)
Limit of (1+n)^3/n^3
Limit of i/(1+i)
Limit of x^2*log(x)
Integral of d{x}
:
x^2-y
The double integral of
:
x^2-y
x^2-y
Identical expressions
x^ two -y
x squared minus y
x to the power of two minus y
x2-y
x²-y
x to the power of 2-y
Similar expressions
e^(-x^2-y^2)*(x^2+y^2)
x^2+y
x^2-y^4
Limit of the function
/
x^2-y
Limit of the function x^2-y
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 - y/ x->oo
$$\lim_{x \to \infty}\left(x^{2} - y\right)$$
Limit(x^2 - y, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(x^{2} - y\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to \infty}\left(x^{2} - y\right)$$ =
$$\lim_{x \to \infty}\left(\frac{1 - \frac{y}{x^{2}}}{\frac{1}{x^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{1 - \frac{y}{x^{2}}}{\frac{1}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{- u^{2} y + 1}{u^{2}}\right)$$
=
$$\frac{- 0^{2} y + 1}{0} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(x^{2} - y\right) = \infty$$
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
Rapid solution
[src]
oo
$$\infty$$
Expand and simplify
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to \infty}\left(x^{2} - y\right) = \infty$$
$$\lim_{x \to 0^-}\left(x^{2} - y\right) = - y$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(x^{2} - y\right) = - y$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(x^{2} - y\right) = - y + 1$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(x^{2} - y\right) = - y + 1$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(x^{2} - y\right) = \infty$$
More at x→-oo