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n^3/(1+n)^3
Limit of the function/ n^3/(1+n)^3

Limit of the function n^3/(1+n)^3

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     /    3   \
     |   n    |
 lim |--------|
n->oo|       3|
     \(1 + n) /
lim⁡n→∞(n3(n+1)3)\lim_{n \to \infty}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) 
Limit(n^3/(1 + n)^3, n, oo, dir='-')
Detail solution
Let's take the limit
lim⁡n→∞(n3(n+1)3)\lim_{n \to \infty}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right)
Let's divide numerator and denominator by n^3:
lim⁡n→∞(n3(n+1)3)\lim_{n \to \infty}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) =
lim⁡n→∞11+3n+3n2+1n3\lim_{n \to \infty} \frac{1}{1 + \frac{3}{n} + \frac{3}{n^{2}} + \frac{1}{n^{3}}}
Do Replacement
u=1nu = \frac{1}{n}
then
lim⁡n→∞11+3n+3n2+1n3=lim⁡u→0+1u3+3u2+3u+1\lim_{n \to \infty} \frac{1}{1 + \frac{3}{n} + \frac{3}{n^{2}} + \frac{1}{n^{3}}} = \lim_{u \to 0^+} \frac{1}{u^{3} + 3 u^{2} + 3 u + 1}
=
103+0⋅3+3⋅02+1=1\frac{1}{0^{3} + 0 \cdot 3 + 3 \cdot 0^{2} + 1} = 1

The final answer:
lim⁡n→∞(n3(n+1)3)=1\lim_{n \to \infty}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = 1
Lopital's rule
We have indeterminateness of type
oo/oo,

i.e. limit for the numerator is
lim⁡n→∞n3=∞\lim_{n \to \infty} n^{3} = \infty
and limit for the denominator is
lim⁡n→∞(n+1)3=∞\lim_{n \to \infty} \left(n + 1\right)^{3} = \infty
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
lim⁡n→∞(n3(n+1)3)\lim_{n \to \infty}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right)
=
lim⁡n→∞(ddnn3ddn(n+1)3)\lim_{n \to \infty}\left(\frac{\frac{d}{d n} n^{3}}{\frac{d}{d n} \left(n + 1\right)^{3}}\right)
=
lim⁡n→∞(n2(n+1)2)\lim_{n \to \infty}\left(\frac{n^{2}}{\left(n + 1\right)^{2}}\right)
=
lim⁡n→∞(ddnn2ddn(n+1)2)\lim_{n \to \infty}\left(\frac{\frac{d}{d n} n^{2}}{\frac{d}{d n} \left(n + 1\right)^{2}}\right)
=
lim⁡n→∞(2n2n+2)\lim_{n \to \infty}\left(\frac{2 n}{2 n + 2}\right)
=
lim⁡n→∞(ddn2nddn(2n+2))\lim_{n \to \infty}\left(\frac{\frac{d}{d n} 2 n}{\frac{d}{d n} \left(2 n + 2\right)}\right)
=
lim⁡n→∞1\lim_{n \to \infty} 1
=
lim⁡n→∞1\lim_{n \to \infty} 1
=
11
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 3 time(s)
The graph
02468-8-6-4-2-1010-20002000
Plot the graph
Rapid solution [src] 
1
11
Expand and simplify
Other limits n→0, -oo, +oo, 1
lim⁡n→∞(n3(n+1)3)=1\lim_{n \to \infty}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = 1
lim⁡n→0−(n3(n+1)3)=0\lim_{n \to 0^-}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = 0
More at n→0 from the left
lim⁡n→0+(n3(n+1)3)=0\lim_{n \to 0^+}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = 0
More at n→0 from the right
lim⁡n→1−(n3(n+1)3)=18\lim_{n \to 1^-}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = \frac{1}{8}
More at n→1 from the left
lim⁡n→1+(n3(n+1)3)=18\lim_{n \to 1^+}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = \frac{1}{8}
More at n→1 from the right
lim⁡n→−∞(n3(n+1)3)=1\lim_{n \to -\infty}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = 1
More at n→-oo
The graph
Limit of the function n^3/(1+n)^3
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