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3/n
Limit of the function/ 3/n

Limit of the function 3/n

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     /3\
 lim |-|
n->oo\n/
lim⁡n→∞(3n)\lim_{n \to \infty}\left(\frac{3}{n}\right) 
Limit(3/n, n, oo, dir='-')
Detail solution
Let's take the limit
lim⁡n→∞(3n)\lim_{n \to \infty}\left(\frac{3}{n}\right)
Let's divide numerator and denominator by n:
lim⁡n→∞(3n)\lim_{n \to \infty}\left(\frac{3}{n}\right) =
lim⁡n→∞(31n1)\lim_{n \to \infty}\left(\frac{3 \frac{1}{n}}{1}\right)
Do Replacement
u=1nu = \frac{1}{n}
then
lim⁡n→∞(31n1)=lim⁡u→0+(3u)\lim_{n \to \infty}\left(\frac{3 \frac{1}{n}}{1}\right) = \lim_{u \to 0^+}\left(3 u\right)
=
0⋅3=00 \cdot 3 = 0

The final answer:
lim⁡n→∞(3n)=0\lim_{n \to \infty}\left(\frac{3}{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
02468-8-6-4-2-1010-5050
Plot the graph
Other limits n→0, -oo, +oo, 1
lim⁡n→∞(3n)=0\lim_{n \to \infty}\left(\frac{3}{n}\right) = 0
lim⁡n→0−(3n)=−∞\lim_{n \to 0^-}\left(\frac{3}{n}\right) = -\infty
More at n→0 from the left
lim⁡n→0+(3n)=∞\lim_{n \to 0^+}\left(\frac{3}{n}\right) = \infty
More at n→0 from the right
lim⁡n→1−(3n)=3\lim_{n \to 1^-}\left(\frac{3}{n}\right) = 3
More at n→1 from the left
lim⁡n→1+(3n)=3\lim_{n \to 1^+}\left(\frac{3}{n}\right) = 3
More at n→1 from the right
lim⁡n→−∞(3n)=0\lim_{n \to -\infty}\left(\frac{3}{n}\right) = 0
More at n→-oo
Rapid solution [src] 
0
00
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The graph
Limit of the function 3/n
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