Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of ((-1+n)/n)^n
-
Limit of (log(x)+sin(x))/(x^2+2*log(x))
-
Limit of -3+3*x+55*x^2/6
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Limit of (9+2*n)/(5+2*n)
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Identical expressions
- ((- one +n)/n)^n
- (( minus 1 plus n) divide by n) to the power of n
- (( minus one plus n) divide by n) to the power of n
- ((-1+n)/n)n
- -1+n/nn
- -1+n/n^n
- ((-1+n) divide by n)^n
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Similar expressions
- ((1+n)/n)^n
- ((-1-n)/n)^n
Limit of the function ((-1+n)/n)^n
The solution
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n
/-1 + n\
lim |------|
n->oo\ n /
$$\lim_{n \to \infty} \left(\frac{n - 1}{n}\right)^{n}$$
Limit(((-1 + n)/n)^n, n, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{n \to \infty} \left(\frac{n - 1}{n}\right)^{n}$$
transform
$$\lim_{n \to \infty} \left(\frac{n - 1}{n}\right)^{n}$$
=
$$\lim_{n \to \infty} \left(\frac{n - 1}{n}\right)^{n}$$
=
$$\lim_{n \to \infty} \left(- \frac{1}{n} + \frac{n}{n}\right)^{n}$$
=
$$\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^{n}$$
=
do replacement
$$u = \frac{n}{-1}$$
then
$$\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^{n}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-1}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-1} = e^{-1}$$
The final answer:
$$\lim_{n \to \infty} \left(\frac{n - 1}{n}\right)^{n} = e^{-1}$$
$$\lim_{n \to \infty} \left(\frac{n - 1}{n}\right)^{n}$$
transform
$$\lim_{n \to \infty} \left(\frac{n - 1}{n}\right)^{n}$$
=
$$\lim_{n \to \infty} \left(\frac{n - 1}{n}\right)^{n}$$
=
$$\lim_{n \to \infty} \left(- \frac{1}{n} + \frac{n}{n}\right)^{n}$$
=
$$\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^{n}$$
=
do replacement
$$u = \frac{n}{-1}$$
then
$$\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^{n}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-1}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-1} = e^{-1}$$
The final answer:
$$\lim_{n \to \infty} \left(\frac{n - 1}{n}\right)^{n} = e^{-1}$$
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
$$\lim_{n \to \infty} \left(\frac{n - 1}{n}\right)^{n} = e^{-1}$$
$$\lim_{n \to 0^-} \left(\frac{n - 1}{n}\right)^{n} = 1$$
More at n→0 from the left
$$\lim_{n \to 0^+} \left(\frac{n - 1}{n}\right)^{n} = 1$$
More at n→0 from the right
$$\lim_{n \to 1^-} \left(\frac{n - 1}{n}\right)^{n} = 0$$
More at n→1 from the left
$$\lim_{n \to 1^+} \left(\frac{n - 1}{n}\right)^{n} = 0$$
More at n→1 from the right
$$\lim_{n \to -\infty} \left(\frac{n - 1}{n}\right)^{n} = e^{-1}$$
More at n→-oo
$$\lim_{n \to 0^-} \left(\frac{n - 1}{n}\right)^{n} = 1$$
More at n→0 from the left
$$\lim_{n \to 0^+} \left(\frac{n - 1}{n}\right)^{n} = 1$$
More at n→0 from the right
$$\lim_{n \to 1^-} \left(\frac{n - 1}{n}\right)^{n} = 0$$
More at n→1 from the left
$$\lim_{n \to 1^+} \left(\frac{n - 1}{n}\right)^{n} = 0$$
More at n→1 from the right
$$\lim_{n \to -\infty} \left(\frac{n - 1}{n}\right)^{n} = e^{-1}$$
More at n→-oo
The graph