Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of (4-x^2)/(3-x^2)
-
Limit of (-2+x^2-x)/(-2+x+3*x^2)
-
Limit of 5-9*x+3*x^2/2
-
Limit of (-16+x^2)/(-64+x^3)
- (x^2+y^2)^2
- Integral of d{x}:
- (x^2+y^2)^2
-
Identical expressions
- (x^ two +y^ two)^ two
- (x squared plus y squared ) squared
- (x to the power of two plus y to the power of two) to the power of two
- (x2+y2)2
- x2+y22
- (x²+y²)²
- (x to the power of 2+y to the power of 2) to the power of 2
- x^2+y^2^2
-
Similar expressions
- (x^2-y^2)^2
Limit of the function (x^2+y^2)^2
The solution
You have entered
[src]
2
/ 2 2\
lim \x + y /
x->oo
$$\lim_{x \to \infty} \left(x^{2} + y^{2}\right)^{2}$$
Limit((x^2 + y^2)^2, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty} \left(x^{2} + y^{2}\right)^{2}$$
Let's divide numerator and denominator by x^4:
$$\lim_{x \to \infty} \left(x^{2} + y^{2}\right)^{2}$$ =
$$\lim_{x \to \infty}\left(\frac{1 + \frac{2 y^{2}}{x^{2}} + \frac{y^{4}}{x^{4}}}{\frac{1}{x^{4}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{1 + \frac{2 y^{2}}{x^{2}} + \frac{y^{4}}{x^{4}}}{\frac{1}{x^{4}}}\right) = \lim_{u \to 0^+}\left(\frac{u^{4} y^{4} + 2 u^{2} y^{2} + 1}{u^{4}}\right)$$
=
$$\frac{0^{4} y^{4} + 2 \cdot 0^{2} y^{2} + 1}{0} = \infty$$
The final answer:
$$\lim_{x \to \infty} \left(x^{2} + y^{2}\right)^{2} = \infty$$
$$\lim_{x \to \infty} \left(x^{2} + y^{2}\right)^{2}$$
Let's divide numerator and denominator by x^4:
$$\lim_{x \to \infty} \left(x^{2} + y^{2}\right)^{2}$$ =
$$\lim_{x \to \infty}\left(\frac{1 + \frac{2 y^{2}}{x^{2}} + \frac{y^{4}}{x^{4}}}{\frac{1}{x^{4}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{1 + \frac{2 y^{2}}{x^{2}} + \frac{y^{4}}{x^{4}}}{\frac{1}{x^{4}}}\right) = \lim_{u \to 0^+}\left(\frac{u^{4} y^{4} + 2 u^{2} y^{2} + 1}{u^{4}}\right)$$
=
$$\frac{0^{4} y^{4} + 2 \cdot 0^{2} y^{2} + 1}{0} = \infty$$
The final answer:
$$\lim_{x \to \infty} \left(x^{2} + y^{2}\right)^{2} = \infty$$
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
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to \infty} \left(x^{2} + y^{2}\right)^{2} = \infty$$
$$\lim_{x \to 0^-} \left(x^{2} + y^{2}\right)^{2} = y^{4}$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(x^{2} + y^{2}\right)^{2} = y^{4}$$
More at x→0 from the right
$$\lim_{x \to 1^-} \left(x^{2} + y^{2}\right)^{2} = y^{4} + 2 y^{2} + 1$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(x^{2} + y^{2}\right)^{2} = y^{4} + 2 y^{2} + 1$$
More at x→1 from the right
$$\lim_{x \to -\infty} \left(x^{2} + y^{2}\right)^{2} = \infty$$
More at x→-oo
$$\lim_{x \to 0^-} \left(x^{2} + y^{2}\right)^{2} = y^{4}$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(x^{2} + y^{2}\right)^{2} = y^{4}$$
More at x→0 from the right
$$\lim_{x \to 1^-} \left(x^{2} + y^{2}\right)^{2} = y^{4} + 2 y^{2} + 1$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(x^{2} + y^{2}\right)^{2} = y^{4} + 2 y^{2} + 1$$
More at x→1 from the right
$$\lim_{x \to -\infty} \left(x^{2} + y^{2}\right)^{2} = \infty$$
More at x→-oo