Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of (9-x^3)/(1+2*x^3)
-
Limit of (-7+8*x^5)/(-3+2*x^5)
-
Limit of (1-6*x+8*x^2)/(7-3*x+3*x^2)
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Limit of (4+8*n)/(6+n)
- Sum of series:
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(1-1/n)^(n^2)
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Identical expressions
- (one - one /n)^(n^ two)
- (1 minus 1 divide by n) to the power of (n squared )
- (one minus one divide by n) to the power of (n to the power of two)
- (1-1/n)(n2)
- 1-1/nn2
- (1-1/n)^(n²)
- (1-1/n) to the power of (n to the power of 2)
- 1-1/n^n^2
- (1-1 divide by n)^(n^2)
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Similar expressions
- (1+1/n)^(n^2)
Limit of the function (1-1/n)^(n^2)
The solution
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/ 2\
\n /
/ 1\
lim |1 - -|
n->oo\ n/
$$\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^{n^{2}}$$
Limit((1 - 1/n)^(n^2), n, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^{n^{2}}$$
transform
do replacement
$$u = \frac{n}{-1}$$
then
$$\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^{n^{2}}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u^{2}}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u^{2}}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{u}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{u} = e^{u}$$
The final answer:
$$\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^{n^{2}} = 0$$
$$\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^{n^{2}}$$
transform
do replacement
$$u = \frac{n}{-1}$$
then
$$\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^{n^{2}}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u^{2}}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u^{2}}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{u}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{u} = e^{u}$$
The final answer:
$$\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^{n^{2}} = 0$$
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 n→0, -oo, +oo, 1
$$\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^{n^{2}} = 0$$
$$\lim_{n \to 0^-} \left(1 - \frac{1}{n}\right)^{n^{2}} = 1$$
More at n→0 from the left
$$\lim_{n \to 0^+} \left(1 - \frac{1}{n}\right)^{n^{2}} = 1$$
More at n→0 from the right
$$\lim_{n \to 1^-} \left(1 - \frac{1}{n}\right)^{n^{2}} = 0$$
More at n→1 from the left
$$\lim_{n \to 1^+} \left(1 - \frac{1}{n}\right)^{n^{2}} = 0$$
More at n→1 from the right
$$\lim_{n \to -\infty} \left(1 - \frac{1}{n}\right)^{n^{2}} = \infty$$
More at n→-oo
$$\lim_{n \to 0^-} \left(1 - \frac{1}{n}\right)^{n^{2}} = 1$$
More at n→0 from the left
$$\lim_{n \to 0^+} \left(1 - \frac{1}{n}\right)^{n^{2}} = 1$$
More at n→0 from the right
$$\lim_{n \to 1^-} \left(1 - \frac{1}{n}\right)^{n^{2}} = 0$$
More at n→1 from the left
$$\lim_{n \to 1^+} \left(1 - \frac{1}{n}\right)^{n^{2}} = 0$$
More at n→1 from the right
$$\lim_{n \to -\infty} \left(1 - \frac{1}{n}\right)^{n^{2}} = \infty$$
More at n→-oo
The graph