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Limit of the function/ -1/n

Limit of the function -1/n

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     /-1 \
 lim |---|
n->oo\ n /
lim⁡n→∞(−1n)\lim_{n \to \infty}\left(- \frac{1}{n}\right) 
Limit(-1/n, n, oo, dir='-')
Detail solution
Let's take the limit
lim⁡n→∞(−1n)\lim_{n \to \infty}\left(- \frac{1}{n}\right)
Let's divide numerator and denominator by n:
lim⁡n→∞(−1n)\lim_{n \to \infty}\left(- \frac{1}{n}\right) =
lim⁡n→∞((−1)1n1)\lim_{n \to \infty}\left(\frac{\left(-1\right) \frac{1}{n}}{1}\right)
Do Replacement
u=1nu = \frac{1}{n}
then
lim⁡n→∞((−1)1n1)=lim⁡u→0+(−u)\lim_{n \to \infty}\left(\frac{\left(-1\right) \frac{1}{n}}{1}\right) = \lim_{u \to 0^+}\left(- u\right)
=
−0=0- 0 = 0

The final answer:
lim⁡n→∞(−1n)=0\lim_{n \to \infty}\left(- \frac{1}{n}\right) = 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
02468-8-6-4-2-1010-2020
Plot the graph
Rapid solution [src] 
0
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Expand and simplify
Other limits n→0, -oo, +oo, 1
lim⁡n→∞(−1n)=0\lim_{n \to \infty}\left(- \frac{1}{n}\right) = 0
lim⁡n→0−(−1n)=∞\lim_{n \to 0^-}\left(- \frac{1}{n}\right) = \infty
More at n→0 from the left
lim⁡n→0+(−1n)=−∞\lim_{n \to 0^+}\left(- \frac{1}{n}\right) = -\infty
More at n→0 from the right
lim⁡n→1−(−1n)=−1\lim_{n \to 1^-}\left(- \frac{1}{n}\right) = -1
More at n→1 from the left
lim⁡n→1+(−1n)=−1\lim_{n \to 1^+}\left(- \frac{1}{n}\right) = -1
More at n→1 from the right
lim⁡n→−∞(−1n)=0\lim_{n \to -\infty}\left(- \frac{1}{n}\right) = 0
More at n→-oo
The graph
Limit of the function -1/n
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