Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of (-16+x^2-6*x)/(-2+x+x^2)
-
Limit of -1+sqrt(5)-x-2/(sqrt(2)-x)
-
Limit of (e^x-x-cos(x))/(-x+log(1+x))
-
Limit of (-cos(x)^3+cos(x))/(4*x*sin(x))
-
Identical expressions
- z^ two -i
- z squared minus i
- z to the power of two minus i
- z2-i
- z²-i
- z to the power of 2-i
-
Similar expressions
- z^2+i
Limit of the function z^2-i
The solution
You have entered
[src]
/ 2 \ lim \z - I/ z->oo
$$\lim_{z \to \infty}\left(z^{2} - i\right)$$
Limit(z^2 - i, z, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{z \to \infty}\left(z^{2} - i\right)$$
Let's divide numerator and denominator by z^2:
$$\lim_{z \to \infty}\left(z^{2} - i\right)$$ =
$$\lim_{z \to \infty}\left(\frac{1 - \frac{i}{z^{2}}}{\frac{1}{z^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{z}$$
then
$$\lim_{z \to \infty}\left(\frac{1 - \frac{i}{z^{2}}}{\frac{1}{z^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{- i u^{2} + 1}{u^{2}}\right)$$
=
$$\frac{- 0^{2} i + 1}{0} = \infty$$
The final answer:
$$\lim_{z \to \infty}\left(z^{2} - i\right) = \infty$$
$$\lim_{z \to \infty}\left(z^{2} - i\right)$$
Let's divide numerator and denominator by z^2:
$$\lim_{z \to \infty}\left(z^{2} - i\right)$$ =
$$\lim_{z \to \infty}\left(\frac{1 - \frac{i}{z^{2}}}{\frac{1}{z^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{z}$$
then
$$\lim_{z \to \infty}\left(\frac{1 - \frac{i}{z^{2}}}{\frac{1}{z^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{- i u^{2} + 1}{u^{2}}\right)$$
=
$$\frac{- 0^{2} i + 1}{0} = \infty$$
The final answer:
$$\lim_{z \to \infty}\left(z^{2} - i\right) = \infty$$
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
Other limits z→0, -oo, +oo, 1
$$\lim_{z \to \infty}\left(z^{2} - i\right) = \infty$$
$$\lim_{z \to 0^-}\left(z^{2} - i\right) = - i$$
More at z→0 from the left
$$\lim_{z \to 0^+}\left(z^{2} - i\right) = - i$$
More at z→0 from the right
$$\lim_{z \to 1^-}\left(z^{2} - i\right) = 1 - i$$
More at z→1 from the left
$$\lim_{z \to 1^+}\left(z^{2} - i\right) = 1 - i$$
More at z→1 from the right
$$\lim_{z \to -\infty}\left(z^{2} - i\right) = \infty$$
More at z→-oo
$$\lim_{z \to 0^-}\left(z^{2} - i\right) = - i$$
More at z→0 from the left
$$\lim_{z \to 0^+}\left(z^{2} - i\right) = - i$$
More at z→0 from the right
$$\lim_{z \to 1^-}\left(z^{2} - i\right) = 1 - i$$
More at z→1 from the left
$$\lim_{z \to 1^+}\left(z^{2} - i\right) = 1 - i$$
More at z→1 from the right
$$\lim_{z \to -\infty}\left(z^{2} - i\right) = \infty$$
More at z→-oo
The graph