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(1-5/n)^n

Limit of the function (1-5/n)^n

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The solution

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            n
     /    5\ 
 lim |1 - -| 
n->oo\    n/ 
limn(15n)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
limn(15n)n\lim_{n \to \infty} \left(1 - \frac{5}{n}\right)^{n}
transform
do replacement
u=n5u = \frac{n}{-5}
then
limn(15n)n\lim_{n \to \infty} \left(1 - \frac{5}{n}\right)^{n} =
=
limu(1+1u)5u\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 5 u}
=
limu(1+1u)5u\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 5 u}
=
((limu(1+1u)u))5\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-5}
The limit
limu(1+1u)u\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}
is second remarkable limit, is equal to e ~ 2.718281828459045
then
((limu(1+1u)u))5=e5\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-5} = e^{-5}

The final answer:
limn(15n)n=e5\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
02468-8-6-4-2-10100.01.0
Rapid solution [src]
 -5
e  
e5e^{-5}
Other limits n→0, -oo, +oo, 1
limn(15n)n=e5\lim_{n \to \infty} \left(1 - \frac{5}{n}\right)^{n} = e^{-5}
limn0(15n)n=1\lim_{n \to 0^-} \left(1 - \frac{5}{n}\right)^{n} = 1
More at n→0 from the left
limn0+(15n)n=1\lim_{n \to 0^+} \left(1 - \frac{5}{n}\right)^{n} = 1
More at n→0 from the right
limn1(15n)n=4\lim_{n \to 1^-} \left(1 - \frac{5}{n}\right)^{n} = -4
More at n→1 from the left
limn1+(15n)n=4\lim_{n \to 1^+} \left(1 - \frac{5}{n}\right)^{n} = -4
More at n→1 from the right
limn(15n)n=e5\lim_{n \to -\infty} \left(1 - \frac{5}{n}\right)^{n} = e^{-5}
More at n→-oo
The graph
Limit of the function (1-5/n)^n