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n^(-3)
Limit of the function/ n^(-3)

Limit of the function n^(-3)

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The solution

You have entered [src] 
     1 
 lim --
n->oo 3
     n
lim⁡n→∞1n3\lim_{n \to \infty} \frac{1}{n^{3}} 
Limit(n^(-3), n, oo, dir='-')
Detail solution
Let's take the limit
lim⁡n→∞1n3\lim_{n \to \infty} \frac{1}{n^{3}}
Let's divide numerator and denominator by n^3:
lim⁡n→∞1n3\lim_{n \to \infty} \frac{1}{n^{3}} =
lim⁡n→∞(1n3)\lim_{n \to \infty}\left(\frac{1}{n^{3}}\right)
Do Replacement
u=1nu = \frac{1}{n}
then
lim⁡n→∞(1n3)=lim⁡u→0+u3\lim_{n \to \infty}\left(\frac{1}{n^{3}}\right) = \lim_{u \to 0^+} u^{3}
=
03=00^{3} = 0

The final answer:
lim⁡n→∞1n3=0\lim_{n \to \infty} \frac{1}{n^{3}} = 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
02468-8-6-4-2-1010-20002000
Plot the graph
Rapid solution [src] 
0
00
Expand and simplify
Other limits n→0, -oo, +oo, 1
lim⁡n→∞1n3=0\lim_{n \to \infty} \frac{1}{n^{3}} = 0
lim⁡n→0−1n3=−∞\lim_{n \to 0^-} \frac{1}{n^{3}} = -\infty
More at n→0 from the left
lim⁡n→0+1n3=∞\lim_{n \to 0^+} \frac{1}{n^{3}} = \infty
More at n→0 from the right
lim⁡n→1−1n3=1\lim_{n \to 1^-} \frac{1}{n^{3}} = 1
More at n→1 from the left
lim⁡n→1+1n3=1\lim_{n \to 1^+} \frac{1}{n^{3}} = 1
More at n→1 from the right
lim⁡n→−∞1n3=0\lim_{n \to -\infty} \frac{1}{n^{3}} = 0
More at n→-oo
The graph
Limit of the function n^(-3)
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