Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of ((-2+x)/(3+x))^(4-x)
-
Limit of (1+n)^3/n^3
- Limit of i/(1+i)
-
Limit of x^2*log(x)
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Identical expressions
- (one +n)^ three /n^ three
- (1 plus n) cubed divide by n cubed
- (one plus n) to the power of three divide by n to the power of three
- (1+n)3/n3
- 1+n3/n3
- (1+n)³/n³
- (1+n) to the power of 3/n to the power of 3
- 1+n^3/n^3
- (1+n)^3 divide by n^3
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Similar expressions
- (1-n)^3/n^3
-
What you mean?
- (1 + n)^3/(n^3)
- (1 + n)^((3/n)^3)
- (1 + n)^((3/n)^3)
You entered:
(1+n)^3/n^3
What you mean?
Limit of the function (1+n)^3/n^3
The solution
You have entered
[src]
/ 3\
|(1 + n) |
lim |--------|
n->oo| 3 |
\ n /
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right)$$
Limit((1 + n)^3/(n^3), n, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right)$$
Let's divide numerator and denominator by n^3:
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{1 + \frac{3}{n} + \frac{3}{n^{2}} + \frac{1}{n^{3}}}{1}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{1 + \frac{3}{n} + \frac{3}{n^{2}} + \frac{1}{n^{3}}}{1}\right) = \lim_{u \to 0^+}\left(u^{3} + 3 u^{2} + 3 u + 1\right)$$
=
$$0^{3} + 3 \cdot 0 + 3 \cdot 0^{2} + 1 = 1$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right) = 1$$
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right)$$
Let's divide numerator and denominator by n^3:
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{1 + \frac{3}{n} + \frac{3}{n^{2}} + \frac{1}{n^{3}}}{1}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{1 + \frac{3}{n} + \frac{3}{n^{2}} + \frac{1}{n^{3}}}{1}\right) = \lim_{u \to 0^+}\left(u^{3} + 3 u^{2} + 3 u + 1\right)$$
=
$$0^{3} + 3 \cdot 0 + 3 \cdot 0^{2} + 1 = 1$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right) = 1$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{n \to \infty} \left(n + 1\right)^{3} = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty} n^{3} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(n + 1\right)^{3}}{\frac{d}{d n} n^{3}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{2}}{n^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} 3 \left(n + 1\right)^{2}}{\frac{d}{d n} 3 n^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{6 n + 6}{6 n}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(6 n + 6\right)}{\frac{d}{d n} 6 n}\right)$$
=
$$\lim_{n \to \infty} 1$$
=
$$\lim_{n \to \infty} 1$$
=
$$1$$
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
$$\lim_{n \to \infty} \left(n + 1\right)^{3} = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty} n^{3} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(n + 1\right)^{3}}{\frac{d}{d n} n^{3}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{2}}{n^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} 3 \left(n + 1\right)^{2}}{\frac{d}{d n} 3 n^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{6 n + 6}{6 n}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(6 n + 6\right)}{\frac{d}{d n} 6 n}\right)$$
=
$$\lim_{n \to \infty} 1$$
=
$$\lim_{n \to \infty} 1$$
=
$$1$$
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
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right) = 1$$
$$\lim_{n \to 0^-}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right) = -\infty$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right) = \infty$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right) = 8$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right) = 8$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right) = 1$$
More at n→-oo
$$\lim_{n \to 0^-}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right) = -\infty$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right) = \infty$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right) = 8$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right) = 8$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right) = 1$$
More at n→-oo
The graph