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5/n^2
Limit of the function/ 5/n^2

Limit of the function 5/n^2

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     /5 \
 lim |--|
n->oo| 2|
     \n /
lim⁡n→∞(5n2)\lim_{n \to \infty}\left(\frac{5}{n^{2}}\right) 
Limit(5/n^2, n, oo, dir='-')
Detail solution
Let's take the limit
lim⁡n→∞(5n2)\lim_{n \to \infty}\left(\frac{5}{n^{2}}\right)
Let's divide numerator and denominator by n^2:
lim⁡n→∞(5n2)\lim_{n \to \infty}\left(\frac{5}{n^{2}}\right) =
lim⁡n→∞(51n21)\lim_{n \to \infty}\left(\frac{5 \frac{1}{n^{2}}}{1}\right)
Do Replacement
u=1nu = \frac{1}{n}
then
lim⁡n→∞(51n21)=lim⁡u→0+(5u2)\lim_{n \to \infty}\left(\frac{5 \frac{1}{n^{2}}}{1}\right) = \lim_{u \to 0^+}\left(5 u^{2}\right)
=
5⋅02=05 \cdot 0^{2} = 0

The final answer:
lim⁡n→∞(5n2)=0\lim_{n \to \infty}\left(\frac{5}{n^{2}}\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
02468-8-6-4-2-10100500
Plot the graph
Other limits n→0, -oo, +oo, 1
lim⁡n→∞(5n2)=0\lim_{n \to \infty}\left(\frac{5}{n^{2}}\right) = 0
lim⁡n→0−(5n2)=∞\lim_{n \to 0^-}\left(\frac{5}{n^{2}}\right) = \infty
More at n→0 from the left
lim⁡n→0+(5n2)=∞\lim_{n \to 0^+}\left(\frac{5}{n^{2}}\right) = \infty
More at n→0 from the right
lim⁡n→1−(5n2)=5\lim_{n \to 1^-}\left(\frac{5}{n^{2}}\right) = 5
More at n→1 from the left
lim⁡n→1+(5n2)=5\lim_{n \to 1^+}\left(\frac{5}{n^{2}}\right) = 5
More at n→1 from the right
lim⁡n→−∞(5n2)=0\lim_{n \to -\infty}\left(\frac{5}{n^{2}}\right) = 0
More at n→-oo
Rapid solution [src] 
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The graph
Limit of the function 5/n^2
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