Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of (-2+2*x^2+log(x))/(e^x-e)
-
Limit of (-x^2+4*x)/(2-sqrt(x))
-
Limit of ((2+3*x)/(-1+3*x))^(-1+4*x)
-
Limit of (3-x+2*x^2)/(5+x^3-8*x)
- Graphing y =:
- n^3/(1+n)^3
-
Identical expressions
- n^ three /(one +n)^ three
- n cubed divide by (1 plus n) cubed
- n to the power of three divide by (one plus n) to the power of three
- n3/(1+n)3
- n3/1+n3
- n³/(1+n)³
- n to the power of 3/(1+n) to the power of 3
- n^3/1+n^3
- n^3 divide by (1+n)^3
-
Similar expressions
- n^3/(1-n)^3
Limit of the function n^3/(1+n)^3
The solution
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/ 3 \
| n |
lim |--------|
n->oo| 3|
\(1 + n) /
$$\lim_{n \to \infty}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right)$$
Limit(n^3/(1 + n)^3, n, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{n \to \infty}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right)$$
Let's divide numerator and denominator by n^3:
$$\lim_{n \to \infty}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right)$$ =
$$\lim_{n \to \infty} \frac{1}{1 + \frac{3}{n} + \frac{3}{n^{2}} + \frac{1}{n^{3}}}$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty} \frac{1}{1 + \frac{3}{n} + \frac{3}{n^{2}} + \frac{1}{n^{3}}} = \lim_{u \to 0^+} \frac{1}{u^{3} + 3 u^{2} + 3 u + 1}$$
=
$$\frac{1}{0^{3} + 0 \cdot 3 + 3 \cdot 0^{2} + 1} = 1$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = 1$$
$$\lim_{n \to \infty}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right)$$
Let's divide numerator and denominator by n^3:
$$\lim_{n \to \infty}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right)$$ =
$$\lim_{n \to \infty} \frac{1}{1 + \frac{3}{n} + \frac{3}{n^{2}} + \frac{1}{n^{3}}}$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty} \frac{1}{1 + \frac{3}{n} + \frac{3}{n^{2}} + \frac{1}{n^{3}}} = \lim_{u \to 0^+} \frac{1}{u^{3} + 3 u^{2} + 3 u + 1}$$
=
$$\frac{1}{0^{3} + 0 \cdot 3 + 3 \cdot 0^{2} + 1} = 1$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = 1$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{n \to \infty} n^{3} = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty} \left(n + 1\right)^{3} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} n^{3}}{\frac{d}{d n} \left(n + 1\right)^{3}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{n^{2}}{\left(n + 1\right)^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} n^{2}}{\frac{d}{d n} \left(n + 1\right)^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{2 n}{2 n + 2}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} 2 n}{\frac{d}{d n} \left(2 n + 2\right)}\right)$$
=
$$\lim_{n \to \infty} 1$$
=
$$\lim_{n \to \infty} 1$$
=
$$1$$
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)
oo/oo,
i.e. limit for the numerator is
$$\lim_{n \to \infty} n^{3} = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty} \left(n + 1\right)^{3} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} n^{3}}{\frac{d}{d n} \left(n + 1\right)^{3}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{n^{2}}{\left(n + 1\right)^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} n^{2}}{\frac{d}{d n} \left(n + 1\right)^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{2 n}{2 n + 2}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} 2 n}{\frac{d}{d n} \left(2 n + 2\right)}\right)$$
=
$$\lim_{n \to \infty} 1$$
=
$$\lim_{n \to \infty} 1$$
=
$$1$$
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
Other limits n→0, -oo, +oo, 1
$$\lim_{n \to \infty}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = 1$$
$$\lim_{n \to 0^-}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = 0$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = 0$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = \frac{1}{8}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = \frac{1}{8}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = 1$$
More at n→-oo
$$\lim_{n \to 0^-}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = 0$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = 0$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = \frac{1}{8}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = \frac{1}{8}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{n^{3}}{\left(n + 1\right)^{3}}\right) = 1$$
More at n→-oo
The graph