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